Creating a proposition from a truth table using only ~ ⋀ and v












2












$begingroup$


I have to find a simple expression for the third column in the truth table using only the logical connectives I've mention above. There are two questions that are involved here.



Problem 1:



Truth table relating to the first problem:



enter image description here



Part 1



All expressions in which p and q return true.



Row 1:  p ⋀ q

Row 3: ~p ⋀ q

Row 4: ~p ⋀ ~q



Thus the expression that satisfy each outcome is: (p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q)



Simplified Expression: 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(p ⋁ q)

 DeMorgan’s Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)

 Commutative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)

 Associative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ (True ⋁ p)

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ True

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((~p ⋁ p) ⋀ (~p ⋁ q))  ⋀ True

 Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (True ⋀ (~p ⋁ q))  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)


Note: I already knew that the answer was (~p v q) but I had to show how I got there.



Problem 2:



Here is the truth table relating to problem number 2:



enter image description here



Part 2 



Row 2:  p ⋀ ~q

Row 3: ~p ⋀ q



Thus the expression that satisfy each outcome is: (p ⋀ ~q) ⋁ (~p ⋀ q)



Simplified Expression:

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ (~p ⋁ ~q) ⋀ (p ⋁ q) 

Rewrite the expression in a conjunction format.

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q) 

DeMorgan’s Law

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q)


Note: I'm not sure if it's possible to simply further



All I'm asking is for someone to check if the steps that I took are right.



Edit: New proof for number 1 please verify!



(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ (~q ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ T Definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p Definition of and

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p  ⋁ (p ⋀ q) commutative Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ p) ⋀ (~p ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ True ⋀ (~p ⋁ q) definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p ⋁ q definition of and





logic propositional-calculus







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$endgroup$








  • 3




    $begingroup$
    You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
    $endgroup$
    – augurar
    Jun 1 '15 at 19:35






  • 3




    $begingroup$
    The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
    $endgroup$
    – Demosthene
    Jun 1 '15 at 19:38










  • $begingroup$
    @Demosthene Alright thanks for checking.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 19:39










  • $begingroup$
    @Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 19:54










  • $begingroup$
    The new proof looks fine.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 23:43
















2












$begingroup$


I have to find a simple expression for the third column in the truth table using only the logical connectives I've mention above. There are two questions that are involved here.



Problem 1:



Truth table relating to the first problem:



enter image description here



Part 1



All expressions in which p and q return true.



Row 1:  p ⋀ q

Row 3: ~p ⋀ q

Row 4: ~p ⋀ ~q



Thus the expression that satisfy each outcome is: (p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q)



Simplified Expression: 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(p ⋁ q)

 DeMorgan’s Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)

 Commutative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)

 Associative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ (True ⋁ p)

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ True

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((~p ⋁ p) ⋀ (~p ⋁ q))  ⋀ True

 Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (True ⋀ (~p ⋁ q))  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)


Note: I already knew that the answer was (~p v q) but I had to show how I got there.



Problem 2:



Here is the truth table relating to problem number 2:



enter image description here



Part 2 



Row 2:  p ⋀ ~q

Row 3: ~p ⋀ q



Thus the expression that satisfy each outcome is: (p ⋀ ~q) ⋁ (~p ⋀ q)



Simplified Expression:

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ (~p ⋁ ~q) ⋀ (p ⋁ q) 

Rewrite the expression in a conjunction format.

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q) 

DeMorgan’s Law

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q)


Note: I'm not sure if it's possible to simply further



All I'm asking is for someone to check if the steps that I took are right.



Edit: New proof for number 1 please verify!



(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ (~q ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ T Definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p Definition of and

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p  ⋁ (p ⋀ q) commutative Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ p) ⋀ (~p ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ True ⋀ (~p ⋁ q) definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p ⋁ q definition of and





logic propositional-calculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
    $endgroup$
    – augurar
    Jun 1 '15 at 19:35






  • 3




    $begingroup$
    The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
    $endgroup$
    – Demosthene
    Jun 1 '15 at 19:38










  • $begingroup$
    @Demosthene Alright thanks for checking.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 19:39










  • $begingroup$
    @Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 19:54










  • $begingroup$
    The new proof looks fine.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 23:43














2












2








2





$begingroup$


I have to find a simple expression for the third column in the truth table using only the logical connectives I've mention above. There are two questions that are involved here.



Problem 1:



Truth table relating to the first problem:



enter image description here



Part 1



All expressions in which p and q return true.



Row 1:  p ⋀ q

Row 3: ~p ⋀ q

Row 4: ~p ⋀ ~q



Thus the expression that satisfy each outcome is: (p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q)



Simplified Expression: 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(p ⋁ q)

 DeMorgan’s Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)

 Commutative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)

 Associative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ (True ⋁ p)

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ True

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((~p ⋁ p) ⋀ (~p ⋁ q))  ⋀ True

 Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (True ⋀ (~p ⋁ q))  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)


Note: I already knew that the answer was (~p v q) but I had to show how I got there.



Problem 2:



Here is the truth table relating to problem number 2:



enter image description here



Part 2 



Row 2:  p ⋀ ~q

Row 3: ~p ⋀ q



Thus the expression that satisfy each outcome is: (p ⋀ ~q) ⋁ (~p ⋀ q)



Simplified Expression:

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ (~p ⋁ ~q) ⋀ (p ⋁ q) 

Rewrite the expression in a conjunction format.

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q) 

DeMorgan’s Law

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q)


Note: I'm not sure if it's possible to simply further



All I'm asking is for someone to check if the steps that I took are right.



Edit: New proof for number 1 please verify!



(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ (~q ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ T Definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p Definition of and

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p  ⋁ (p ⋀ q) commutative Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ p) ⋀ (~p ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ True ⋀ (~p ⋁ q) definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p ⋁ q definition of and





logic propositional-calculus







share|cite|improve this question











$endgroup$




I have to find a simple expression for the third column in the truth table using only the logical connectives I've mention above. There are two questions that are involved here.



Problem 1:



Truth table relating to the first problem:



enter image description here



Part 1



All expressions in which p and q return true.



Row 1:  p ⋀ q

Row 3: ~p ⋀ q

Row 4: ~p ⋀ ~q



Thus the expression that satisfy each outcome is: (p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q)



Simplified Expression: 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(p ⋁ q)

 DeMorgan’s Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)

 Commutative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)

 Associative Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ (True ⋁ p)

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((p ⋀ q) ⋁ ~p) ⋀ True

 Definition of ⋁

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ((~p ⋁ p) ⋀ (~p ⋁ q))  ⋀ True

 Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (True ⋀ (~p ⋁ q))  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)  ⋀ True

 P ⋀ True ≡ P

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ q)


Note: I already knew that the answer was (~p v q) but I had to show how I got there.



Problem 2:



Here is the truth table relating to problem number 2:



enter image description here



Part 2 



Row 2:  p ⋀ ~q

Row 3: ~p ⋀ q



Thus the expression that satisfy each outcome is: (p ⋀ ~q) ⋁ (~p ⋀ q)



Simplified Expression:

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ (~p ⋁ ~q) ⋀ (p ⋁ q) 

Rewrite the expression in a conjunction format.

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q) 

DeMorgan’s Law

(p ⋀ ~q) ⋁ (~p ⋀ q)  ≡ ~(p ⋀ q) ⋀ (p ⋁ q)


Note: I'm not sure if it's possible to simply further



All I'm asking is for someone to check if the steps that I took are right.



Edit: New proof for number 1 please verify!



(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ (~q ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p ⋀ T Definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (p ⋀ q) ⋁ ~p Definition of and

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p  ⋁ (p ⋀ q) commutative Law 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ (~p ⋁ p) ⋀ (~p ⋁ q) Distributive Law

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ True ⋀ (~p ⋁ q) definition of v 

(p ⋀ q) ⋁ (~p ⋀ q) ⋁ (~p ⋀ ~q) ≡ ~p ⋁ q definition of and





logic propositional-calculus




logic propositional-calculus






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edited Jun 1 '15 at 21:47







Deathslice

















asked Jun 1 '15 at 19:31









DeathsliceDeathslice

12619




12619








  • 3




    $begingroup$
    You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
    $endgroup$
    – augurar
    Jun 1 '15 at 19:35






  • 3




    $begingroup$
    The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
    $endgroup$
    – Demosthene
    Jun 1 '15 at 19:38










  • $begingroup$
    @Demosthene Alright thanks for checking.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 19:39










  • $begingroup$
    @Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 19:54










  • $begingroup$
    The new proof looks fine.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 23:43














  • 3




    $begingroup$
    You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
    $endgroup$
    – augurar
    Jun 1 '15 at 19:35






  • 3




    $begingroup$
    The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
    $endgroup$
    – Demosthene
    Jun 1 '15 at 19:38










  • $begingroup$
    @Demosthene Alright thanks for checking.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 19:39










  • $begingroup$
    @Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 19:54










  • $begingroup$
    The new proof looks fine.
    $endgroup$
    – Doug Spoonwood
    Jun 1 '15 at 23:43








3




3




$begingroup$
You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
$endgroup$
– augurar
Jun 1 '15 at 19:35




$begingroup$
You may want to look up Karnaugh maps, a good way to find simplified forms of boolean expressions up to 4 variables.
$endgroup$
– augurar
Jun 1 '15 at 19:35




3




3




$begingroup$
The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
$endgroup$
– Demosthene
Jun 1 '15 at 19:38




$begingroup$
The first one is correct. For the second one, $(plandneg q)lor(neg pland q)$ is already the most succinct expression. Note that this is in fact $neg(pleftrightarrow q)$.
$endgroup$
– Demosthene
Jun 1 '15 at 19:38












$begingroup$
@Demosthene Alright thanks for checking.
$endgroup$
– Deathslice
Jun 1 '15 at 19:39




$begingroup$
@Demosthene Alright thanks for checking.
$endgroup$
– Deathslice
Jun 1 '15 at 19:39












$begingroup$
@Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
$endgroup$
– Doug Spoonwood
Jun 1 '15 at 19:54




$begingroup$
@Demosthene The first one is most definitely not correct. The author says that he/she gets from "(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)" to "((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)" by an associative law.
$endgroup$
– Doug Spoonwood
Jun 1 '15 at 19:54












$begingroup$
The new proof looks fine.
$endgroup$
– Doug Spoonwood
Jun 1 '15 at 23:43




$begingroup$
The new proof looks fine.
$endgroup$
– Doug Spoonwood
Jun 1 '15 at 23:43










1 Answer
1






active

oldest

votes


















0












$begingroup$

A way to check that such steps are correct can get understood by writing out the specific use of the logical laws you've used, and then highlighting in some way via underlining or circling where you made the replacement. I'll provide an example.



I use Polish notation. So, your first problem starts with



A A Kpq KNpq K NpNq.



You use the De Morgan's law which says that



K NxNy = NA x y in your first step.



Thus, we'll highlight the relevant part of the relevant formula.



A A Kpq KNpq K NpNq.



And substituting in De Morgan's law we have



K NpNq = NA p q.



Now making the replacement we see that



A A Kpq KNpq NA p q is correct.



In your problem you say that



"(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)"



leads to



"((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)"



via the associative law (for ⋁?).



I don't follow. Which part did you replace? To me it looks like the first part you replaced was the parentheses with a blank. You replaced



"(~p ⋀ q) ⋁ ~(q ⋁ p)" with "~p) ⋀ ((q ⋁ ~q) ⋁ p)". I'm very sure that no law allows you to do that.



You went from



" ((p ⋀ q) ⋁ ~p) ⋀ True"



to



"((~p ⋁ p) ⋀ (~p ⋁ q)) ⋀ True"



You can do that, but I encourage you to look at the relevant distributive law more closely. I suspect you'll find it doesn't allow you to do exactly what you did here... you need another law invoked also.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:03










  • $begingroup$
    In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:05










  • $begingroup$
    If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:22










  • $begingroup$
    How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:02










  • $begingroup$
    I make a new proof for number 1 can you please check?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:22











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$begingroup$

A way to check that such steps are correct can get understood by writing out the specific use of the logical laws you've used, and then highlighting in some way via underlining or circling where you made the replacement. I'll provide an example.



I use Polish notation. So, your first problem starts with



A A Kpq KNpq K NpNq.



You use the De Morgan's law which says that



K NxNy = NA x y in your first step.



Thus, we'll highlight the relevant part of the relevant formula.



A A Kpq KNpq K NpNq.



And substituting in De Morgan's law we have



K NpNq = NA p q.



Now making the replacement we see that



A A Kpq KNpq NA p q is correct.



In your problem you say that



"(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)"



leads to



"((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)"



via the associative law (for ⋁?).



I don't follow. Which part did you replace? To me it looks like the first part you replaced was the parentheses with a blank. You replaced



"(~p ⋀ q) ⋁ ~(q ⋁ p)" with "~p) ⋀ ((q ⋁ ~q) ⋁ p)". I'm very sure that no law allows you to do that.



You went from



" ((p ⋀ q) ⋁ ~p) ⋀ True"



to



"((~p ⋁ p) ⋀ (~p ⋁ q)) ⋀ True"



You can do that, but I encourage you to look at the relevant distributive law more closely. I suspect you'll find it doesn't allow you to do exactly what you did here... you need another law invoked also.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:03










  • $begingroup$
    In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:05










  • $begingroup$
    If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:22










  • $begingroup$
    How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:02










  • $begingroup$
    I make a new proof for number 1 can you please check?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:22
















0












$begingroup$

A way to check that such steps are correct can get understood by writing out the specific use of the logical laws you've used, and then highlighting in some way via underlining or circling where you made the replacement. I'll provide an example.



I use Polish notation. So, your first problem starts with



A A Kpq KNpq K NpNq.



You use the De Morgan's law which says that



K NxNy = NA x y in your first step.



Thus, we'll highlight the relevant part of the relevant formula.



A A Kpq KNpq K NpNq.



And substituting in De Morgan's law we have



K NpNq = NA p q.



Now making the replacement we see that



A A Kpq KNpq NA p q is correct.



In your problem you say that



"(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)"



leads to



"((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)"



via the associative law (for ⋁?).



I don't follow. Which part did you replace? To me it looks like the first part you replaced was the parentheses with a blank. You replaced



"(~p ⋀ q) ⋁ ~(q ⋁ p)" with "~p) ⋀ ((q ⋁ ~q) ⋁ p)". I'm very sure that no law allows you to do that.



You went from



" ((p ⋀ q) ⋁ ~p) ⋀ True"



to



"((~p ⋁ p) ⋀ (~p ⋁ q)) ⋀ True"



You can do that, but I encourage you to look at the relevant distributive law more closely. I suspect you'll find it doesn't allow you to do exactly what you did here... you need another law invoked also.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:03










  • $begingroup$
    In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:05










  • $begingroup$
    If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:22










  • $begingroup$
    How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:02










  • $begingroup$
    I make a new proof for number 1 can you please check?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:22














0












0








0





$begingroup$

A way to check that such steps are correct can get understood by writing out the specific use of the logical laws you've used, and then highlighting in some way via underlining or circling where you made the replacement. I'll provide an example.



I use Polish notation. So, your first problem starts with



A A Kpq KNpq K NpNq.



You use the De Morgan's law which says that



K NxNy = NA x y in your first step.



Thus, we'll highlight the relevant part of the relevant formula.



A A Kpq KNpq K NpNq.



And substituting in De Morgan's law we have



K NpNq = NA p q.



Now making the replacement we see that



A A Kpq KNpq NA p q is correct.



In your problem you say that



"(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)"



leads to



"((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)"



via the associative law (for ⋁?).



I don't follow. Which part did you replace? To me it looks like the first part you replaced was the parentheses with a blank. You replaced



"(~p ⋀ q) ⋁ ~(q ⋁ p)" with "~p) ⋀ ((q ⋁ ~q) ⋁ p)". I'm very sure that no law allows you to do that.



You went from



" ((p ⋀ q) ⋁ ~p) ⋀ True"



to



"((~p ⋁ p) ⋀ (~p ⋁ q)) ⋀ True"



You can do that, but I encourage you to look at the relevant distributive law more closely. I suspect you'll find it doesn't allow you to do exactly what you did here... you need another law invoked also.






share|cite|improve this answer









$endgroup$



A way to check that such steps are correct can get understood by writing out the specific use of the logical laws you've used, and then highlighting in some way via underlining or circling where you made the replacement. I'll provide an example.



I use Polish notation. So, your first problem starts with



A A Kpq KNpq K NpNq.



You use the De Morgan's law which says that



K NxNy = NA x y in your first step.



Thus, we'll highlight the relevant part of the relevant formula.



A A Kpq KNpq K NpNq.



And substituting in De Morgan's law we have



K NpNq = NA p q.



Now making the replacement we see that



A A Kpq KNpq NA p q is correct.



In your problem you say that



"(p ⋀ q) ⋁ (~p ⋀ q) ⋁ ~(q ⋁ p)"



leads to



"((p ⋀ q) ⋁ ~p) ⋀ ((q ⋁ ~q) ⋁ p)"



via the associative law (for ⋁?).



I don't follow. Which part did you replace? To me it looks like the first part you replaced was the parentheses with a blank. You replaced



"(~p ⋀ q) ⋁ ~(q ⋁ p)" with "~p) ⋀ ((q ⋁ ~q) ⋁ p)". I'm very sure that no law allows you to do that.



You went from



" ((p ⋀ q) ⋁ ~p) ⋀ True"



to



"((~p ⋁ p) ⋀ (~p ⋁ q)) ⋀ True"



You can do that, but I encourage you to look at the relevant distributive law more closely. I suspect you'll find it doesn't allow you to do exactly what you did here... you need another law invoked also.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 1 '15 at 19:52









Doug SpoonwoodDoug Spoonwood

8,08212244




8,08212244












  • $begingroup$
    Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:03










  • $begingroup$
    In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:05










  • $begingroup$
    If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:22










  • $begingroup$
    How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:02










  • $begingroup$
    I make a new proof for number 1 can you please check?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:22


















  • $begingroup$
    Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:03










  • $begingroup$
    In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:05










  • $begingroup$
    If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 20:22










  • $begingroup$
    How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:02










  • $begingroup$
    I make a new proof for number 1 can you please check?
    $endgroup$
    – Deathslice
    Jun 1 '15 at 21:22
















$begingroup$
Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
$endgroup$
– Deathslice
Jun 1 '15 at 20:03




$begingroup$
Thanks for the suggestions. What I normally do is to check if both sides remain logically equivalent. In regards to what I did when it came to the associative law, that is where I was having a slight doubt of what I was doing. I thought that maybe having the same logical operator would allow me to perform the associative law and then do what I did after that.
$endgroup$
– Deathslice
Jun 1 '15 at 20:03












$begingroup$
In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
$endgroup$
– Deathslice
Jun 1 '15 at 20:05




$begingroup$
In any case, the last expression that I got for the problem is right because it satisfy each condition in the truth table. Therefore, (~p v q) is logically equivalent to p implies q.
$endgroup$
– Deathslice
Jun 1 '15 at 20:05












$begingroup$
If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
$endgroup$
– Deathslice
Jun 1 '15 at 20:22




$begingroup$
If I rewrite the expressions like this (~(~p v ~q) v ~(p v ~q)) v ~(p v q), could I then use the associative law?
$endgroup$
– Deathslice
Jun 1 '15 at 20:22












$begingroup$
How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
$endgroup$
– Deathslice
Jun 1 '15 at 21:02




$begingroup$
How would I use the associative law for something like this: ~(~q ⋁ ~p) ⋁ ~p ⋀ T
$endgroup$
– Deathslice
Jun 1 '15 at 21:02












$begingroup$
I make a new proof for number 1 can you please check?
$endgroup$
– Deathslice
Jun 1 '15 at 21:22




$begingroup$
I make a new proof for number 1 can you please check?
$endgroup$
– Deathslice
Jun 1 '15 at 21:22


















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