Why is $((p \land q) \Rightarrow z) \Rightarrow (p \Rightarrow z) \lor (q \Rightarrow z)$ true?

Question

I will propose a counterexample to $$((p \land q) \Rightarrow z) \Rightarrow ((p \Rightarrow z) \lor (q \Rightarrow z)).$$ Let's assume that $p$ is "$n$ is divisible by $2$", $q$ is "$n$ is divisible by $5$" and $z$ is "$n$ divisible by $10$". Then, while it's true that ($n$ is divisible by $2$ and $n$ is divisible by $5$) is sufficient for ($n$ is divisible by $10$), the disjunction ($n$ is divisible by $2$ is sufficient for $n$ is divisible by $10$) OR ($n$ is divisible by $5$ is sufficient for $n$ is divisible by $10$) is actually false.

What is wrong with my above counterexample and understanding? Why is $$((p \land q) \Rightarrow z) \Rightarrow ((p \Rightarrow z) \lor (q \Rightarrow z))$$ true?

Answer 1

The sentences \begin{align}\Big((P \land Q) → Z\Big) → \Big((P → Z) \lor (Q→ Z)\Big)\tag{✔️1}\end{align} and \begin{align}∀\color{green}n\bigg(\quad\Big(\big(P\color{green}n ∧ Q\color{green}n\big) → Z\color{green}n\Big) \;→\; \Big(\;\big(P\color{green}n → Z\color{green}n\big) \;∨\; \big(Q\color{green}n → Z\color{green}n\big)\;\Big)\quad\bigg)\tag{✔️2}\end{align} are both indeed true regardless of interpretation (use a truth table to check this for Sentence $1$), because for any given value of $\color{green}n,$ if $P$ and $Q$ jointly implies $Z$ then at least one of $P$ and $Q$ implies $Z.$

On the other hand, your counterexample

while it's true that ($\color\red n$ is divisible by $2$ and $\color\red n$ is divisible by $5$) is sufficient for ($\color\red n$ is divisible by $10$), the disjunction ($\color{blue} n$ is divisible by $2$ is sufficient for $\color{blue} n$ is divisible by $10$) OR ($\color\violet n$ is divisible by $5$ is sufficient for $\color\violet n$ is divisible by $10$) is actually false

disproves the logical validity of \begin{align}∀\color{red}n\Big(\big(P\color{red}n ∧ Q\color{red}n\big) → Z\color{red}n\Big) \;→\; \bigg(∀\color{blue}n\big(P\color{blue}n → Z\color{blue}n\big) \;∨\; ∀\color{violet}n\big(Q\color{violet}n → Z\color{violet}n\big)\bigg),\tag{❌3}\end{align} which is more clearly written as \begin{align}∀\color\red a\Big(\big(P\color\red a ∧ Q\color\red a\big) → Z\color\red a\Big) \;→\; \bigg(∀\color{blue}b\big(P\color{blue}b → Z\color{blue}b\big) \;∨\; ∀\color{violet}c\big(Q\color{violet}c → Z\color{violet}c\big)\bigg).\tag{❌3}\end{align} Your counterexample isn't disproving Sentence $2,$ because it isn't premised on its seven occurrences of $n$ having a common value.

Answer 2

First things first: Draw out a truth table, always helpful for understanding this type of question!

I believe the confusion comes from the difference between how we use the implication sign in common language and what it means as a connective in propositional calculus: I can't add an image of the truth table , but I would be grateful if someone would.

If either p or q were False, then the right hand side is True. So the expression is True.

So we only need to consider both p and q True, then if z false, the LHS is True (As False--> False is True), then the implication is True( As False-->True and False--> False is True so we don't need to look at the RHS).

If both p and q True, and z True, then the LHS is True, and the RHS is clearly true, so the implication is True.. So under any possible truth assignment the statement is true. That's known as a tautology!

Answer 3

I will first present a speedy proof of this, and then an explanation as to why this works that I hope is more clear than the others that have been given.

Assume that $((p \land q) \implies z) \implies ((p \implies z) \lor (q \implies z))$ is false. Then it must be the case that If $p \land q \implies z$ is true, it can be that $p \implies z \lor q \implies z $ is false, meaning $$(p \land q \implies z) \land \neg ((p \implies z) \lor (q \implies z))\\ (\neg(p \land q) \lor z) \land ((p \land \neg z) \land (q \land \neg z))$$ This is forced to be a contradiction. By the second parenthesis term, p and q are true and z is false, but by the right hand side if $z$ is false the $p$ or $q$ is false which is a contradiction, and if $z$ is true then it contradicts that $z$ is true. Therefore, the statement cannot be false and is a tautology. $\square$

Now, let's talk about what the sentence means and why your example doesn't make sense. I think the issue here is your prepositions don't exactly match up with what the sentence is written as. You would like to say that for a certain number $n$, if $2|n \wedge 5|n \implies 10|n$, which it does, then either $2|n \implies 10|n$ or $5|n \implies 10|n$. These are not causally true statements, but they are logically true. If we let $n=7$ the statement is clearly vacuously true. If $n=10$, you showed that it holds already. But what if $n = 15$? Then, $2 |15$ is false, $5 | 15$ is true, and $10 | 15$ is false. But since $2|15 \implies 10|15$ is vacuously true, the statement $$(2|n \implies 10|n) \lor (5|n \implies 10|n) \equiv (T) \lor (5|n\implies 10|n) \equiv T$$ Still evaluates to true. So in my own very loose words, this statement is true by vacuous truth mainly, and when its not equivalent it is just equivalent to $p \land q \implies z$.

This is a really good question however, as it does raise an excellent point to something most people don't realize, which is when and where vacuous truth applies, and why, even though many people say it just makes sense, it is very often confusing.