Proving a statement of the form $Prightarrow(lnot Qlor lnot R)$.
Not looking for a proof of my question, only an answer to my question below.
Question: I want to assume that $P$ is true for this proof, then if I assume that $Q$ is also true and conclude that $R$ is not true, does that prove this statement? Do I also have to prove that if $R$ is true then $Q$ is not? Thank you for the help.
For reference the logical form of the statement: Suppose $f:[a,b]rightarrowBbb{R}$ is not constant on $[a,b]$, then $f$ is not continuous or $f([a,b])⊄Bbb{Q}^c$.
real-analysis logic proof-theory
edited Nov 27 at 19:58
Shaun
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
add a comment |
Not looking for a proof of my question, only an answer to my question below.
Question: I want to assume that $P$ is true for this proof, then if I assume that $Q$ is also true and conclude that $R$ is not true, does that prove this statement? Do I also have to prove that if $R$ is true then $Q$ is not? Thank you for the help.
For reference the logical form of the statement: Suppose $f:[a,b]rightarrowBbb{R}$ is not constant on $[a,b]$, then $f$ is not continuous or $f([a,b])⊄Bbb{Q}^c$.
real-analysis logic proof-theory
edited Nov 27 at 19:58
Shaun
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13
add a comment |
Not looking for a proof of my question, only an answer to my question below.
Question: I want to assume that $P$ is true for this proof, then if I assume that $Q$ is also true and conclude that $R$ is not true, does that prove this statement? Do I also have to prove that if $R$ is true then $Q$ is not? Thank you for the help.
For reference the logical form of the statement: Suppose $f:[a,b]rightarrowBbb{R}$ is not constant on $[a,b]$, then $f$ is not continuous or $f([a,b])⊄Bbb{Q}^c$.
real-analysis logic proof-theory
edited Nov 27 at 19:58
Shaun
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
Not looking for a proof of my question, only an answer to my question below.
Question: I want to assume that $P$ is true for this proof, then if I assume that $Q$ is also true and conclude that $R$ is not true, does that prove this statement? Do I also have to prove that if $R$ is true then $Q$ is not? Thank you for the help.
For reference the logical form of the statement: Suppose $f:[a,b]rightarrowBbb{R}$ is not constant on $[a,b]$, then $f$ is not continuous or $f([a,b])⊄Bbb{Q}^c$.
real-analysis logic proof-theory
real-analysis logic proof-theory
edited Nov 27 at 19:58
Shaun
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
edited Nov 27 at 19:58
Shaun
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
edited Nov 27 at 19:58
Shaun
8,647113680
edited Nov 27 at 19:58
Shaun
8,647113680
edited Nov 27 at 19:58
Shaun
8,647113680
8,647113680
asked Nov 27 at 1:47
Albert Diaz
925
asked Nov 27 at 1:47
Albert Diaz
925
asked Nov 27 at 1:47
Albert Diaz
925
925
You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13
add a comment |
You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13
You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13
You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
You're right.
This is because $lnot Qlorlnot R$ is equivalent to $Qto lnot R$. There is no need to prove $Rtolnot Q$ if you prove the latter (due to the symmetry of $lor$).
Also: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$
answered Nov 27 at 2:02
Shaun
8,647113680
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
add a comment |
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1 Answer
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1 Answer
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oldest
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You're right.
This is because $lnot Qlorlnot R$ is equivalent to $Qto lnot R$. There is no need to prove $Rtolnot Q$ if you prove the latter (due to the symmetry of $lor$).
Also: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$
answered Nov 27 at 2:02
Shaun
8,647113680
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
add a comment |
You're right.
This is because $lnot Qlorlnot R$ is equivalent to $Qto lnot R$. There is no need to prove $Rtolnot Q$ if you prove the latter (due to the symmetry of $lor$).
Also: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$
answered Nov 27 at 2:02
Shaun
8,647113680
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
add a comment |
You're right.
This is because $lnot Qlorlnot R$ is equivalent to $Qto lnot R$. There is no need to prove $Rtolnot Q$ if you prove the latter (due to the symmetry of $lor$).
Also: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$
answered Nov 27 at 2:02
Shaun
8,647113680
You're right.
This is because $lnot Qlorlnot R$ is equivalent to $Qto lnot R$. There is no need to prove $Rtolnot Q$ if you prove the latter (due to the symmetry of $lor$).
Also: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$
answered Nov 27 at 2:02
Shaun
8,647113680
edited Nov 27 at 19:57
answered Nov 27 at 2:02
Shaun
8,647113680
answered Nov 27 at 2:02
Shaun
8,647113680
answered Nov 27 at 2:02
Shaun
8,647113680
8,647113680
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
add a comment |
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
1
1
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 at 2:03
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 at 2:09
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 at 2:10
add a comment |
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You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 at 2:13