Proving a statement of the form $Prightarrow(lnot Qlor lnot R)$.












0














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







share|cite|improve this question
























  • You might be better off proving the contrapositive of the statement . . .
    – Shaun
    Nov 27 '18 at 2:13
















0














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







share|cite|improve this question
























  • You might be better off proving the contrapositive of the statement . . .
    – Shaun
    Nov 27 '18 at 2:13














0












0








0


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:58









Shaun

8,760113680




8,760113680










asked Nov 27 '18 at 1:47









Albert Diaz

925




925












  • You might be better off proving the contrapositive of the statement . . .
    – Shaun
    Nov 27 '18 at 2:13


















  • You might be better off proving the contrapositive of the statement . . .
    – Shaun
    Nov 27 '18 at 2:13
















You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 '18 at 2:13




You might be better off proving the contrapositive of the statement . . .
– Shaun
Nov 27 '18 at 2:13










1 Answer
1






active

oldest

votes


















1














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: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$






share|cite|improve this answer



















  • 1




    Thank you, logical statements can be confusing at times for me but now I recall that symmetry
    – Albert Diaz
    Nov 27 '18 at 2:03










  • You're welcome, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:09










  • I've added to the answer, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:10











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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

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1














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: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$






share|cite|improve this answer



















  • 1




    Thank you, logical statements can be confusing at times for me but now I recall that symmetry
    – Albert Diaz
    Nov 27 '18 at 2:03










  • You're welcome, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:09










  • I've added to the answer, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:10
















1














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: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$






share|cite|improve this answer



















  • 1




    Thank you, logical statements can be confusing at times for me but now I recall that symmetry
    – Albert Diaz
    Nov 27 '18 at 2:03










  • You're welcome, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:09










  • I've added to the answer, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:10














1












1








1






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: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$






share|cite|improve this answer














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: $$begin{align}
Pto(lnot Qlorlnot R)&equiv (lnot P)lor((lnot Q)lor(lnot R))\ &equiv ((lnot P)lor(lnot Q))lor(lnot R)\ &equiv (lnot(Pland Q))lorlnot R \ &equiv (Pland Q)to lnot R.end{align}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 19:57

























answered Nov 27 '18 at 2:02









Shaun

8,760113680




8,760113680








  • 1




    Thank you, logical statements can be confusing at times for me but now I recall that symmetry
    – Albert Diaz
    Nov 27 '18 at 2:03










  • You're welcome, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:09










  • I've added to the answer, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:10














  • 1




    Thank you, logical statements can be confusing at times for me but now I recall that symmetry
    – Albert Diaz
    Nov 27 '18 at 2:03










  • You're welcome, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 at 2:09










  • I've added to the answer, @AlbertDiaz :)
    – Shaun
    Nov 27 '18 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 '18 at 2:03




Thank you, logical statements can be confusing at times for me but now I recall that symmetry
– Albert Diaz
Nov 27 '18 at 2:03












You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 '18 at 2:09




You're welcome, @AlbertDiaz :)
– Shaun
Nov 27 '18 at 2:09












I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 '18 at 2:10




I've added to the answer, @AlbertDiaz :)
– Shaun
Nov 27 '18 at 2:10


















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