# truth tables and validity of arguments

1. $p$
2. $p \to q$
3. $\lnot q \lor r$
4. $\therefore r$

In order to prove validity with truth tables, do 1) 2) and 3) have to be true in order for the conclusion to be true?

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Try omitting on of the first three propositions, and see if you can still derive the conclusion – David H Feb 24 '14 at 17:53

A valid argument is an argument IF the premises are true, then the conclusion must be true. That is, for an argument to be valid, we must have that the truth of the premises guarantee/entail the truth of the conclusion.

An argument can be valid even if one or more of the premises are false. Provided the premises logically entail the conclusion, the argument is valid. If we know the premises to be true, in a valid argument, we then know the conclusion is true. If we don't know the truth values of the premises, we can show such an argument is valid if we can show that, "if the premises are/ were true, then the conclusion is necessarily true."

Whether or not the premises are true is another question: for example, a sound argument requires that the argument be valid and that the premises are all true.

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Roy wants to show validity of argument $p$. Then, what he has to show is that "if the premises are/ were true, then the conclusion is necessarily true". The cases of $p$ being true while some of the premises might not be, are out of question, right? – frabala Feb 24 '14 at 18:07
@frabala - what does it mean "validity of argument $p$" ? A formula is valid (in prop calculus a tautology) and $p$ is not (no single prop letter is) or an argument is, in which case (as per OP's answer) must be something like : "from A and B and .., X follows". – Mauro ALLEGRANZA Feb 25 '14 at 12:36
This answer doesn't talk about proving validity with truth tables. – Doug Spoonwood Feb 26 '14 at 17:17
Hey, @Doug, do you frequent Philosophy.SE? If so, is logic a tag there? Do you see more logic there, or here? – amWhy Feb 26 '14 at 17:32
@amWhy I've been there a few times. I haven't gone there recently though. There is a logic tag there. There's currently 426 questions there tagged logic at philosophy (beta), and it's the tag with the most number of questions. There's 3,954 questions tagged logic here. Some questions at philosophy.stackexchange I'd expect would come as off-topic here. – Doug Spoonwood Feb 26 '14 at 17:42

The problem is an easy one and you do not necessarly have to "write down" all the truth table.

In order to prove with truth tables that

$\Gamma \vDash r$

where $\Gamma = \{p, p \rightarrow q, \lnot q \lor r \}$ we apply the definition :

$\Gamma$ tautologically implies $\alpha$ (written $\Gamma \vDash \alpha$) iff every truth assignment for the sentence symbols in $\Gamma$ and $\alpha$ that satisfies every member of $\Gamma$ also satisfies $\alpha$.

In this case, we can "play a little bit" with the definition. We rephrase it in the equivalent form :

there is no truth assignment for the sentence symbols in $\Gamma$ and $\alpha$ that satisfies every member of $\Gamma$ and does not satisfy $\alpha$.

Proving this with truth-tables amount to showing that there is no rows in the tt where all the formulas in $\Gamma$ are evaluated to true and the formula $r$ is evaluated to false.

We can take benefit of the fact that $\lnot q \lor r$ is equivalent to $q \rightarrow r$.

Let we start: we assume a truth assignment $v$ such that, for all $\varphi \in \Gamma$, $v(\varphi)=true$.

We have $v(p)=true$, and by truth-functional property of $\rightarrow$, also $v(q)=true$, in order $p \rightarrow q$ to be true.

So $v(r)=true$, for the same reason (we have assumed that, for the truth assignment $v$, $\lnot q \lor r$, i.e. $q \rightarrow r$, is true).

In conclusion, the truth assignment $v$, such that all the premises (the formulas in $\Gamma$) are true, boils down to the row of the tt corresponding to : t, t, t.

For this truth assignment $v$, $r$ is true; but $v$ is a generic truth assignment that satisfies all the premises, so we cannot have a truth assignment that satisfy all the premises and that evaluates $r$ to false.

In conclusion, we have proved that :

$\{p, p \rightarrow q, \lnot q \lor r \} \vDash r$.

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This is exactly what I meant in my answer. Ok, I have to admit I should be more analytic. :) – frabala Feb 25 '14 at 14:26

Yes, by definition $\Gamma\vDash p$, if $p$ is true whenever $\forall \gamma\in\Gamma$, $\gamma$ is true.

Here, $\Gamma$ contains (1), (2) and (3).

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An argument can be valid even if one or more of the premises are false. Provided the premises logically entail the conclusion, the argument is valid. If we know the premises to be true, in a valid argument we then know the conclusion is true. If we don't know the truth values of the premises, but can show such an argument is valid if we can show that, "if the premises are/ were true, then the conclusion is necessarily true." – amWhy Feb 24 '14 at 17:54
@amWhy I think this is the same, because $(A_1\wedge_\cdots\wedge A_n)\to B$ is logically equivalent to $A_1\to(A_2\to\cdots\to(A_n\to B)$. I don't see were is your disagreement, here. – frabala Feb 24 '14 at 18:00
But note an implication requires the following: "IF" the antecedent is true, then the conclusion is true. An implication holds when every the antecedent (premises) is false, as well. Logical entailment is even stronger. It requires that IF the premises are true, then so must be the conclusion... We classify an argument as valid or not valid strictly on the basis of whether the conclusion necessarily follows from the premise, even when we know nothing about the truth values of the premises. $$(p \rightarrow q),\;p\;$$ guarantees $q$, regardless of whether p, q are true or what they mean. – amWhy Feb 24 '14 at 18:09
@amWhy I still think we're saying the same thing. $(p\to q), p\vdash q$ holds, because whenever $p$ is true and $(p\to q)$ is true, $q$ is also true. – frabala Feb 24 '14 at 18:18
@amWhy no he is saying quite the opposite, the question is about "when is an argument valid?" while this answer answers "what is a valid argument". Hope he changes his answer otherwise I will need to vote it down (and i don't like that) – Willemien Feb 25 '14 at 7:42

No, 1., 2., and 3. do not have to be true in order to prove validity with truth tables.

You can prove the valid argument

(p $\land$ $\lnot$ p)

q

using truth tables as follows:

  p   q   ¬p  (p ∧ ¬ p)  q
F   F   T      F       F
F   T   T      F       T
T   F   F      F       F
T   T   F      F       T


Consequently, you could prove the following argument:

1. (p $\land$ $\lnot$ p)
2. r
3. s

q

with truth tables.

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The question calls for a look at the truth table proving the validity of the argument set out, so here it is.

It reveals that propositions 1) and 2) in the original argument can be false when the conclusion "r" is true. So in those cases the answer to the question is no.

Interestingly though, when proposition 3) is false, so is r. So 3) has to be true for r to be true.

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