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  1. p ∨ q is not tautology.
  2. p ∨ q is tautology.
  3. p ∧ q contradiction.
  4. (p → q) → (q → p) is tautology.
  5. All of the above are wrong.

I tried to use a truth table and I got (1) but apparently $q=f$ and $p=f$ cannot be in the truth table so the answer is not (1). Can anyone explain why one is not the answer? and also, what is the correct answer?

Harish Chandra Rajpoot
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Dror
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  • $F\lor F\ne T $ – John Douma Jun 29 '22 at 18:47
  • Welcome to Math.SE! I have tried to improve the readability of your question by improving the Markdown code. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. – GNUSupporter 8964民主女神 地下教會 Jun 29 '22 at 19:09
  • Is this question from a quiz? I don't like it because it ostensibly requires the candidate to reason from the two given premises yet, as I was saying in its recent duplicate target here and evidenced by Manx's answer below, the premises turn out to be redundant, in which case the candidate is left wondering whether they are required to use the two premises in their argument. – ryang Jun 30 '22 at 03:18

1 Answers1

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Let $a,b$ be sentence letters.

$1.$ Is it true that for any well formed formula $p,q$ say if $\not\vDash p\to q$ and $\not\vDash q\to p$ then $\not\vDash p \lor q?$

Consider $p:\equiv a$ and $q:\equiv\lnot a$ which gives $\not\vDash a\to \lnot a$ and $\not\vDash \lnot a\to a$ but $\vDash a \lor \lnot a$.

$2.$ Is it true that for any well formed formula $p,q$ say if $\not\vDash p\to q$ and $\not\vDash q\to p$ then $\vDash p \lor q?$

Consider $p:\equiv a$ and $q:\equiv b$ which gives $\not\vDash a\to b$ and $\not\vDash b\to a$ but $\not\vDash a \lor b$.

$3.$ Is it true that for any well formed formula $p,q$ say if $\not\vDash p\to q$ and $\not\vDash q\to p$ then $\vDash \lnot p \lor\lnot q?$

Consider $p:\equiv a$ and $q:\equiv b$ which gives $\not\vDash a\to b$ and $\not\vDash b\to a$ but $\not\vDash \lnot a \lor \lnot b$.

$4.$ For any well formed formula $p,q$ say if $\not\vDash p\to q$ and $\not\vDash q\to p$ then $\vDash (p\to q)\to(q\to p)?$

Note that $(p\to q)\to(q\to p)\equiv p\lor\lnot q$.

Consider $p:\equiv a$ and $q:\equiv b$ which gives $\not\vDash a\to b$ and $\not\vDash b\to a$ but $\not\vDash a \lor \lnot b$.

Hence the right answer is

$5.$ All of the above are wrong.

Ethan
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  • Out of curoisty, what does the double turnstile mean? I know that the regular turnstil means, "from $A$ I know $B$", $A \vdash B$. Tried looking on the internet about it, but didn't find much right off the bat. – James Jun 30 '22 at 02:06
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    @Michael $A\vdash B$ is the syntactic consequence means there is a deduction from $A$ to $B$. $A\vDash B$ is the semantic consequence means in every interpretation that $A$ is true we have $B$ is also true, and $\vDash A$ just means $A$ is true in every interpretation i.e. $A$ is a tautology. – Ethan Jun 30 '22 at 02:48