I'm trying to prove P→(Q∧R) from the following premises: (P→Q)∧R, (P∧R)→S, ¬S [on hold]
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
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put on hold as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom 2 days ago
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
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Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
New contributor
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
New contributor
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 5:24
Orlando Piedrahita
11
New contributor
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 5:24
Orlando Piedrahita
11
New contributor
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 16 at 5:24
Orlando Piedrahita
11
asked Nov 16 at 5:24
Orlando Piedrahita
11
11
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Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Orlando Piedrahita is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
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Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
add a comment |
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
answered Nov 16 at 6:07
Fefnir Wilhelm
62
62
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fefnir Wilhelm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
add a comment |
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
2 days ago
add a comment |
