Does $ P(x,y,z) \wedge x \Rightarrow P(T,y,z) $ in logic?

Question

I am studying the logic section of discrete mathematics. While thinking about a proof problem, I wanted to use $$ (p \rightarrow (\lnot q \lor r)) \land q $$ to prove $$ p \rightarrow r$$ I assumed that since $q$ is $T$, then $\lnot q$ is $F$, and $F \lor r \equiv r$, so I naively simplified the original expression to $$p \rightarrow (F \lor r)$$ and then concluded that $p \rightarrow r$.

I realized that I was wrong because no one said that directly substituting truth values into an expression during a proof is valid, and it seemed to be a coincidence.

However, I constructed several other expressions where this substitution method was valid. Is it reasonable to do so, or is it purely coincidental?

Answer 1

Does $ P(\color\red x,y,z) \wedge \color\red x \Rightarrow P(\color\red T,y,z) $ in logic?

Suppose that a compound proposition $P$ contains proposition $\color\red X,$ and proposition $Q$ is $P$ but with all instances of $\color\red X$ replaced by a tautology. You're asking whether the conjunction of $P$ and $\color\red X$ entails $Q.$ In other words: is it legitimate to assume a hypothesis to be true?

The answer is: Yes! In particular:

I naively simplified $$ (p \rightarrow \lnot q \lor r) \land q \tag1$$ to $$p \rightarrow F \lor r\tag2$$ I realized that I was wrong

(1) does entail (2), so your "simplification" is actually a valid deduction.

I put quotation marks around "simplification" because, of course, (1) and (2) aren't equivalent to each other: put $$p\;:\Leftrightarrow\;r\;:\Leftrightarrow\;(1=1)\\q\;:\Leftrightarrow\;(1\ne1).$$ Similarly, for $\color\red{X}\;:\Leftrightarrow\;(1\ne1),$ $$\big(\color\red{\boldsymbol\top}\lor(1\ne1)\big)\kern.6em\not\kern-.6em\implies\big(\color\red{X}\lor(1\ne1)\big)\land \color\red{X}.$$

Answer 2

What you are doing is very closely related to the general Biconditional Substitution equivalence principle, which states:

$$F(\phi) \land (\phi \leftrightarrow \psi) \Leftrightarrow F(\psi) \land (\phi \leftrightarrow \psi)$$

where $F(\psi)$ is the result of replacing any, some, or all instances of $\phi$ that occur as part of the formula $F(\phi)$ with $\psi$.

So in your case, you can go through the following equivalences:

$$ (p \rightarrow (\lnot q \lor r)) \land q $$

$$\Leftrightarrow \text{(Identity)}$$

$$ (p \rightarrow (\lnot q \lor r)) \land (q \leftrightarrow \top)$$

$$\Leftrightarrow \text{(Biconditional Substitution)}$$

$$(p \rightarrow (\lnot \top \lor r)) \land (q \leftrightarrow \top)$$

$$\Leftrightarrow \text{(Inverse)}$$

$$(p \rightarrow (\bot \lor r)) \land (q \leftrightarrow \top)$$

$$\Leftrightarrow \text{(Identity)}$$

$$(p \rightarrow r) \land (q \leftrightarrow \top)$$

$$\Leftrightarrow \text{(Identity)}$$

$$(p \rightarrow r) \land q$$

and from that last statement, you can now infer $p \rightarrow r$, thus ultimately showing that $ (p \rightarrow (\lnot q \lor r)) \land q $ logically implies $p \rightarrow r$

Bonus: Biconditonal substitution is a super useful principle when solving Knights and Knaves puzzles. Consider the following Knights and Knaves Puzzles (as found here):

A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet two inhabitants: Zoey and Mel. Zoey tells you that Mel is a knave. Mel says, “Neither Zoey nor I are knaves.” Can you determine who is a knight and who is a knave?

OK, let's symbolize: Zoey is a knights if and only if what Zoey says is true, and so we get:

$Z \leftrightarrow \neg M$

Likewise for Mel:

$M \leftrightarrow \neg (Z \lor M)$

OK, so let's put that together and simplify:

$$(Z \leftrightarrow \neg M) \land (M \leftrightarrow \neg (Z \lor M))$$

$$\Leftrightarrow \text{(Biconditional Substitution)}$$

$$(Z \leftrightarrow \neg M) \land (M \leftrightarrow \neg (\neg M \lor M))$$

$$\Leftrightarrow \text{(Complement)}$$

$$(Z \leftrightarrow \neg M) \land (M \leftrightarrow \neg \top)$$

$$\Leftrightarrow \text{(Inverse)}$$

$$(Z \leftrightarrow \neg M) \land (M \leftrightarrow \bot)$$

$$\Leftrightarrow \text{(Biconditional Substitution)}$$

$$(Z \leftrightarrow \neg \bot) \land (M \leftrightarrow \bot)$$

$$\Leftrightarrow \text{(Inverse)}$$

$$(Z \leftrightarrow \top) \land (M \leftrightarrow \bot)$$

$$\Leftrightarrow \text{(Identity)}$$

$$Z \land \neg M$$

Answer 3

You can't arbitrarily assign truth values to propositions because you only know how they are being used together with other propositions.

However, what you did can be formalized in some way. You said,

I assumed that since $q$ is $T$, then $\lnot q$ is $F$, and $F \lor r \equiv r$, so I naively simplified the original expression to $$p \rightarrow (F \lor r)$$ and then concluded that $p \rightarrow r$.

This is called disjunctive syllogism. That is, $P \land (\lnot P \lor Q) \Rightarrow Q$. It's a tautology, so regardless of the truth values of $P$ and $Q$, it would still be true.