Missteps in deriving a conclusion from premises

Question

The problem is to show that $q \rightarrow r$ is a conclusion of the premises $$(p\wedge t) \rightarrow (r \vee s),\\q \rightarrow (u \wedge t),\\u \rightarrow p,\\\neg s.$$

Why is the following answer false?

Assume that $q$ is true.

1. By the modus ponens, we conclude that $u \wedge t$ is valid.
2. By simplification rule, $u$ and $t$ are valid, respectively.
3. $p \wedge t$ is valid by conjunction
4. By modus ponens, $r\vee s$ is valid.
5. $r$ is valid by disjunctive syllogism.
6. We conclude $q \rightarrow r$ is valid since $q$ and $r$ are valid.

I think that 1) and 6) are problematic. If so, why?

Do we need to consider the case $q$=False?

Answer 1

No, 1. and 6. are not the main problem, but they are not correct like all other propositions due to one error in all of them: you used the word "valid" instead of "true". Validity in Logic is something completely different than being true, in fact, it only really applies to arguments, not to propositions. An argument being valid means that if all the premises of the argument are true, the conclusion cannot be false. Therefore, you should rewrite every instance of "valid" in your proof with "true".

You don't need to suppose that $q$ is false, since to prove a conditional, all you need to do is assume the left side is true and with that arrive to the conclusion that the right side is true, which is exactly what you did.

However your main problem is the lack of one step:

2.5. $p$ is true by modus ponens because we have $u \rightarrow p$ and $u$ is true.

Because 2. doesn't follow automatically to 3. since you never explicitly showed that $p$ is true.

Answer 2

The exercise is to derive the conclusion $(q\to r)$ from the given premises.

1. Do we need to consider the case $q$=False?

   Your correct strategy appears to be assuming (additionally to the four premises $\phi$) that $q$ is true, then finally deriving that $r$ is also true: $$\phi\to(q\to r)\equiv (\phi\land q)\to r.$$

   You don't need to consider the $q$=False case, because the rules of inference rest on the premises (including your fifth premise $q$) being true; after all, from a false premise, by the principle of explosion, we can conclude anything we wish!

2. Alternatively, by deriving either that $q$ is false or that $r$ is true, you can then conclude (in either case) that $q\to r$ must be true, since \begin{array}{cc|c@{}ccc@{}c} q&r&(&q&\rightarrow&r&)\\\hline 1&1&&1&\mathbf{1}&1&\\ 1&0&&1&\mathbf{0}&0&\\ 0&1&&0&\mathbf{1}&1&\\ 0&0&&0&\mathbf{1}&0&. \end{array}

3. The critical error in your derivation lies in step #3, where you appear to be assuming that the antecedent $(p\wedge t)$ of the premise $(p\wedge t)\rightarrow (r \vee s)$ is true, rather than that the entire premise is true; remember, $(A\to B)$ means that if $A$ is true, then $B$ is true; $(A\to B)$ can be true while $A$ is false.

   Instead, just make use of the unused premise $u\to p$ to justify your step #3.

4. You've been wrongly conflating the terms “valid” and “true”. Every single instance of “valid” needs to be replaced with “true”.

   A propositional-logic sentence being valid means that it is logically true, which means that it is true whichever the truth values of its atomic sentences. For example, a proposition containing the atomic sentences $A,B,C$ being valid means that the proposition is true whichever of the eight combination of values $(T,T,T),(T,T,F),(T,F,F),$ etc. the triple $(A,B,C)$ takes on, otherwise the proposition is invalid.

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EDIT

Thank you for the answer. I added the following step: since $u$ is true, and $u\rightarrow p$ is true, we can get $p$ is true by the modus ponens. Then, is the answer still wrong?

Then your derivation becomes largely correct.

Here are a few more remarks: modus ponens is a name so doesn't require “the”, steps #1,4,5 ought to be elaborated on (like how you wrote your inserted step), and step #0 ought to be revised to “assume that the above premises and $q$ are all true.”