# Is it a tautology or not? According to my truth table its not

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.

• What is "contingent"? Neither a tautology nor a contradiction? – Yuval Filmus May 25 '19 at 15:27
• Is there any relation between $A,B,C$? – Yuval Filmus May 25 '19 at 15:27
• 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." – logicsnewbie2019 May 25 '19 at 16:07
• What does "(insert sentence)" mean? Do you have particular sentences in mind? Perhaps you should just describe the question as stated. – Yuval Filmus May 25 '19 at 16:09
• if [(q<->p)<->s] is a tautology and p->s is contingent, does it follow that q->s) is contingent? – 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?

Equivalently,

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.
• 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!! – logicsnewbie2019 May 25 '19 at 16:36
• I don't think you understand the question very well. I suggest reviewing course material, or contacting the lecturer or a TA. – Yuval Filmus May 25 '19 at 16:37
• true, thats why i was asking here since its weekend and no one is working :) – logicsnewbie2019 May 25 '19 at 16:53
• 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. – 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.