First of all, I only learn "logic" in Real Analysis. So please bear with me and be welcome to define "logic" if necessary.

In First-order logic, "If P then Q" can be expressed as "(not P) or Q". Then the Truth table will tell us that if statement P is False, then the whole statement will be True regardless whether Q is true or not. I understand this, however, when it comes to the "logic" behind the implication. It confuses me.

So my question is

The assumption P is False, but we Suppose it is TRUE (if part). Even the logic behind the implication is wrong, the statement "If P then Q" is still true?

I need to emphasise that we want to assume P is TRUE though we know it is FALSE

For example, back to the time when aircraft cannot fly. Someone said, "If the aircraft can fly, then people inside it would die due to the high speed when it takes off. "

Using First-order logic, since the aircraft at that time cannot fly, so the whle statement (if-then) will be true. However, the "logic" in this statement seems to be wrong. We admit that the aircraft at that time cannot fly at that time, but we want to consider the situation where it can (maybe in the future or even just a physics thought experiment). Under this assumption, people don't die due to high speed of the aircraft. Then we would conclude that this is a False statement.

Then it will arise a natural questions like,

what should we do if we want to make inference or implication under some conditions which might not be true? Does the "logic" just lost the control after a False assumption?

How should we make a statement to distinguish "True because the logic is true" and "True because the assumption is False"

Let me rephrase my question:

Is there any theory focusing on the logic in implication instead of just true and false of the whole statement?

so that even the assumption is false, if the logic of implication does not hold then we will consider the proof/implication if False.

For example, a question on exam paper,

  • "Prove $\sqrt 2 \notin \mathbf{Q}$".

If a student answers, "suppose $\sqrt 2 \in \mathbf{Q}$, then $\pi \in \mathbf{Q}$. Contradiction."

This answer technically is a True statement and indeed we arrived in a contradiction. But $\pi \in \mathbf{Q}$ does not follow from $\sqrt 2 \in \mathbf{Q}$ in logical sense. So it deserves 0 mark.

An example that is built on a false antecedent but implication is valid:

  • If (1+1=1, 2+2=2^2, 3+3=3^2, etc), then n+n=n^2.

The antecedent is clearly False, but suppose instead it is true, say we redefine + as x, then logic follows. While for "if ... then n+n=0", the implication doesn't follow any "logic".

  • $\begingroup$ I think you might be describing the concept of vacuous truth: en.wikipedia.org/wiki/Vacuous_truth $\endgroup$ – 1729 Jul 3 '17 at 19:38
  • $\begingroup$ It's hard to parse your question, but it might be related to vacuous truth and/or principle of explosion. $\endgroup$ – user223391 Jul 3 '17 at 19:43
  • $\begingroup$ In math a statement does not change its truth value with time: $2+2=4$ is true today; was true yesterday and will be true tomorrow. $\endgroup$ – Mauro ALLEGRANZA Jul 3 '17 at 19:54
  • $\begingroup$ If you want to consider possibility, you have to use modal logic. $\endgroup$ – Mauro ALLEGRANZA Jul 3 '17 at 19:55
  • $\begingroup$ @TomHolt yeah, it is a related concept. However, I am confused about the "logic" of the implication under a False/Fake/Imaginary condition. There are some cases where antecedent can be wrong and right, so "neutral" in general, like it is raining today; I ate an apple, etc, under this assumption, which means we takes a step back, suppose it is true, then ***. Can't we say it is wrong if the consequent is out of logic? $\endgroup$ – tautology Jul 3 '17 at 20:07

If $P$ then $Q$ means if $P$ is true, then, no matter what, $Q$ is true.

Hence you cannot have $P$ true and $Q$ false.

So either $Q$ is true or, if $Q$ is false, since you cannot have $P$ true then $P$ is false, which is $\neg P$ true. Which you can rewrite as $$\neg P \vee Q $$

If the lights are red then cars stop. Whatever the cars are doing when the lights are NOT red will not change the truth of this statement.

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  • $\begingroup$ Thx for your answer but it is not what I am after $\endgroup$ – tautology Jul 3 '17 at 20:12
  • $\begingroup$ @Furrane Is "If P is true, then Q is true" is the same as "If P, then Q" why is "is true" usually omitted? Thanks in advance. $\endgroup$ – user599310 Apr 7 at 22:18

Some comments

1) About: "(not P) or Q" (in symbols: $\lnot P \lor Q$), if we use the truth-table definition for the connectives, we agree to evaluate it as True when at least one of the disjuncts is True.

Thus, if we know that $P$ is False, by excluded middle, we know also that $\lnot P$ is True, and thus we can conclude that the disjunction is True regardless of the truth value of $Q$.

But we want so: if we agree to model "or" in a non-exclusive way, we agree on the fact that one is enough.

2) About: "I need to emphasise that we want to assume P is TRUE though we know it is FALSE", this is not clear.

It can be useful to avoid to conflate the antecedent $P$ of a conditional $P \to Q$ with an assumption of an argument.

In a conditional, we have no "assumptions"; by bivalence, either $P$ is true or $P$ is false (in this case the "or" is in fact exclusive).

If $P$ is False, then by truth-table for conditional, $P \to Q$ is True.

This fact is another side of the truth-functional definition of "or"; $\lnot P \lor Q$ is (in classical logic) equivalent to $P \to Q$, and this equivalence is "built in" into the truth table definitions of the connectives.

3) A possible source of misunderstanding is in your example about:

"If a student answers, 'suppose $\sqrt 2 \in \mathbb Q$, then $\pi \in \mathbb Q$. Contradiction'. This is a true conditional but a wrong poof."

Correct, but for the "wrong reason".

The conditional: "if $\sqrt 2 \in \mathbb Q$, then $\pi \in \mathbb Q$" is True, but the exam ask you to "Prove that $\sqrt 2 \notin \mathbb Q$".

The proof must be an argument showing that, from true premises and using correct inference steps, we are able to derive the conclusion: $\sqrt 2 \notin \mathbb Q$.

One useful kind of argument is the proof by contradiction: we assume the negation of the sought conclusion.

In this case we assume: $\sqrt 2 \in \mathbb Q$ and we derive a contradicition, i.e. the negation of some already proved fact, like - as in the case of the well-known example - that a certain number is both even and odd.

Thus, applying the "pattern" of the proof by contradicition, we conclude that the assumption (i.e. $\sqrt 2 \in \mathbb Q$) is untenable, and thus (by excluded middle) we are licensed by the pattern of the proof to assert the negation of the assumption.

The negation of the assumption is:

$\sqrt 2 \notin \mathbb Q$

and we conclude the proof.

Conclusion: if we have proved a contradiction from assumption $P$ (i.e. if we have derived from it a false mathematical fact), then we can discard the assumption as false and (by excluded middle) assert its negation: $\lnot P$.

Going back again to the example, if we have proved that:

"supposing $\sqrt 2 \in \mathbb Q$, it follws that a number is both even and odd (i.e. a contradiction)",

we can conclude with the falsity of our supposition, i.e. that $\sqrt 2 \notin \mathbb Q$.

4) Is there some link between the two facts: conditional and proof ? YES.

Let see why. We have a purported fact to be proved: call it $\lnot P$ (i.e. not-($\sqrt 2 \in \mathbb Q$)).

We have a contradicition; call it False.

We have an argument according to the pattern:

assume $P$ and derive False; then conclude with $\lnot P$.

Consider the conditional: "if $P$, then False", i.e. $P \to \text{ False }$.

If we have proved it (we have proved that: assuming $P$, we can derive "logically" False) we have established a mathematical fact and thus we are entitled to assert that the conditional $P \to \text{ False }$ is True.

But, according to the truth table for the conditional, there is only one possibility left for the conditional $P \to \text{ False }$ to be True: is when the antecedent is False.

This means that $P$ is False, and thus $\lnot P$ must be True.

Conclusion: if we have proved a contradiction from assumption $P$ (i.e. if we have derived from it a false mathematical fact), then we can discard the assumption as false and (by excluded middle) assert its negation: $\lnot P$.

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  • $\begingroup$ Thanks for your clear explanation. Probably I did not use a proper example to show what I want to ask. The example of square root 2 is not about "how proof by contradiction works" or "why if the antecedent is False then the conditional is True". I am simply wondering if a conditional is True, but the logic inside the conditional is False, how should we call it? Like, can we say the conditional is "logically False"? In other word, is there a type of "Truth Value" for "logic in a conditional/implication". $\endgroup$ – tautology Jul 4 '17 at 17:46
  • $\begingroup$ It happens in many cases that a statement is True, but just does not make sense in logic. It also happens when the antecedent P is true (At first I restrict the case when it is not). For example, "If duck is an animal, then any individual will die." This statement is also true, but no logic shows in the implication. So we might need a truth value for the logic inside the statement, which in this case is "False". $\endgroup$ – tautology Jul 4 '17 at 17:57
  • $\begingroup$ @tautology - what does it mean: " the logic inside the conditional" ??? $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 19:21
  • $\begingroup$ well, I don't know how to phrase it correctly. But I think you shall understand from the example, duck is an animal does not imply any individual will die. The statement is true, but the logic in the statement is shaky $\endgroup$ – tautology Jul 4 '17 at 19:53
  • $\begingroup$ @tautology - I've discussed it already in my answer the difference between "if P, then Q" and "from P we can prove (derive) Q". A correct mathematical proof can derive from the assumption $x > 0$ the conclusion $x \ge 0$. A correct mathematical prof can derive from the assumption $0=1$ the conclusion $0=0$. $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 20:05

An appendix to the observations in the answer by @MauroALLEGRANZA. The sentence

$$\sqrt{2} \in \mathbb{Q} \rightarrow \pi \in \mathbb{Q}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$

is not a tautology, unlike, say,

$$(\sqrt{2} \in \mathbb{Q} \wedge \pi \in \mathbb{Q}) \rightarrow \pi \in \mathbb{Q} \enspace.~~~~~~~(2)$$

Rather, $(1)$ is a statement that is true of the real numbers under the standard interpretation of the non-logical symbols that appear in it. (Those symbols are $\sqrt{\phantom{1}}$, $2$, $\in$, $\mathbb{Q}$, and $\pi$.) If, in a fit of extravagance, we decided that $\sqrt{\phantom{1}}$ should henceforth denote the identity function, the sentence would become false.

Not so for $(2)$, because all sentences of the form $(p \wedge q) \rightarrow q$ are always true by virtue of their logical structure---they are tautologies. Changing the interpretation of $\sqrt{\phantom{1}}$, or of any other non-logical symbol, has no effect on the truth value of $(2)$.

The upshot of $(1)$ not being a tautology is that to use it in a proof as a stepping stone, one needs to prove it first.

If an enterprising student produced $(1)$ as an alleged proof of the irrationality of $\sqrt{2}$ and later complained of receiving little or no credit, a reasonable course of action would be to ask, "Please, show me more of your proof. Specifically, show me your proof that $(1)$ is true."

If that "proof" were, "Since $\sqrt{2}$ is irrational the antecedent is false; hence the implication is true," we would be facing a petitio principii: the assumption of the irrationality of $\sqrt{2}$ in a proof of the irrationality of $\sqrt{2}$.

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