If $\bigr((q\leftrightarrow p)\leftrightarrow s\bigl)$ is a tautology and $p\rightarrow s$ is contingent, does it follow that $q\rightarrow s$ is contingent?

Since I can't show $\bigr((q\leftrightarrow p)\leftrightarrow s\bigl)$ is a tautology, I'm unsure how to proceed.

  • $\begingroup$ What is "contingent"? Neither a tautology nor a contradiction? $\endgroup$ – Yuval Filmus May 25 '19 at 15:27
  • $\begingroup$ Is there any relation between $A,B,C$? $\endgroup$ – Yuval Filmus May 25 '19 at 15:27
  • $\begingroup$ correct, contingent is neither a tautology or contradiction. no relation other than what you can gather from "if A is a tautology (insert sentence), and B (insert sentence) is contingent, does it follow that C (insert sentence) is contingent." $\endgroup$ – logicsnewbie2019 May 25 '19 at 16:07
  • $\begingroup$ What does "(insert sentence)" mean? Do you have particular sentences in mind? Perhaps you should just describe the question as stated. $\endgroup$ – Yuval Filmus May 25 '19 at 16:09
  • $\begingroup$ if [(q<->p)<->s] is a tautology and p->s is contingent, does it follow that q->s) is contingent? $\endgroup$ – logicsnewbie2019 May 25 '19 at 16:13

The question is not phrased clearly. Perhaps the question can be rephrased as follows: Let $p,q$ and $s$ be propositions whose truth values belong to some subset $A \subseteq \{T,F\}^3$. Suppose $A$ is such that $(q \leftrightarrow p) \leftrightarrow s$ evaluates to true for each assignment from $A$ and $p \rightarrow s$ can take both possible truth values over assignments from $A$. Can $q \rightarrow s$ take both possible truth values over assignments from $A$?

But there are many $A$'s which satisfy the given two conditions, and different $A$'s will give different answers- it can be a contingency, or it can be tautology.


Let me start by conjecturing a more complete version of your question:

Is it true that for any Boolean expressions $q,p,s$, if $(q \leftrightarrow p) \leftrightarrow s$ is a tautology and $p \to s$ is contingent, then necessarily $q \to s$ is contingent?


Is it true that for any subset $B$ of truth assignments for $(q,p,s)$, if $(q \leftrightarrow p) \leftrightarrow s$ is a tautology with respect to $B$ (i.e., every truth assignment in $B$ satisfies it) and $p \to s$ is contingent with respect to $B$ (i.e., some truth assignment in $B$ satisfies it, and another one falsifies it), then $q \to s$ is also contingent with respect to $B$?

The answer is negative: it doesn't follow that $q \to s$ is contingent. Suppose that $q = \bot$, $p = \lnot s$, and $s$ is a variable (i.e., its truth value is arbitrary). Then:

  • $(q \leftrightarrow p) \leftrightarrow s$ is always true.
  • $p \to s$ is true iff $s$ is true.
  • $q \to s$ is always true.
  • $\begingroup$ am so new to logics that im not following why you have to suppose things. Any chance you can explain in a different way? i very much appreciate your help!! $\endgroup$ – logicsnewbie2019 May 25 '19 at 16:36
  • $\begingroup$ I don't think you understand the question very well. I suggest reviewing course material, or contacting the lecturer or a TA. $\endgroup$ – Yuval Filmus May 25 '19 at 16:37
  • $\begingroup$ true, thats why i was asking here since its weekend and no one is working :) $\endgroup$ – logicsnewbie2019 May 25 '19 at 16:53
  • $\begingroup$ Unfortunately I can't help you any further since I don't have the background (indeed, have no idea where does this question belong) and don't want to confuse you; though my answer does seem valid. $\endgroup$ – Yuval Filmus May 25 '19 at 17:38

$\bigr((q\leftrightarrow p)\leftrightarrow s\bigl)$ is not a tautology since setting $q=true$ $p=true$ and $s=false$ falsifies it. Your hypotheisis is then $false$. Since $false$ implies anything, you can conclude anything you want. However,if you replace $\bigr((q\leftrightarrow p)\leftrightarrow s\bigl)$ by a formula that is indeed a tautology,then you can simply use a truth table to check the conclusion.


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