I study intuitive logic and I often come across implications. I noticed that sometimes the statement $p \implies q$ seems to mean completely different things: when I write a proof, I am allowed to assume $p$ to prove $q$. In other words, I am allowed to assume that the hypothesis is true. But when I apply a method of mathematical induction, for example, I need to prove the hypothesis in order to show that the conclusion in true. In other words, I am not allowed to assume that the hypothesis is true. Hence my question: do we consider logical implication only depending on the context? In other words, when you see the statement of the form $p \implies q$, is it true that you are not allowed to assume that $p$ is true if you look at this statement out of context?
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Are you speaking about propositions which are only vacuously true? Like "all primes between $8$ and $10$ are divisible by $103$"? Usually for induction you want to prove an abstract statement of the form "$P(n)\implies P(n+1)$" without referring to the truth or falseness of $P(n)$, and then to separately establish that some instance, like maybe $P(1)$, is actually true. – lulu Nov 02 '24 at 16:43
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3What you're calling "context" is the logical structure of whatever mathematical argument is being made. Just "seeing" a proposition doesn't make anything true. If you assume $p$ and deduce $q$, then you have proved $p\implies q$ using conditional proof. – Karl Nov 02 '24 at 16:51
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1Induction has a double conditional form: if P(0) is true and if for every k, if P(k) then P(k+1) is true, then P(n) is true for every n. Thus, in order to conclude that P(n) holds for every n, we ha to prove (not assume) the first two conditions. – Mauro ALLEGRANZA Nov 02 '24 at 17:00
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1Can you explain what you mean by "when I apply a method of mathematical induction, for example, I need to prove the hypothesis in order to show that the conclusion in true." ? – Ted Nov 02 '24 at 18:12
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To instructors: Maybe more basic methods of proof should be taught before induction, e.g. how to structure a simple proof with a premise and a conclusion, how to generalize, etc.? Maybe learn to walk before you can run? Start with proving something like: A & B => B & A. – Dan Christensen Nov 08 '24 at 15:15
2 Answers
...when I write a proof, I am allowed to assume p to prove q.
Yes. Then you could conclude that $p \implies q.$ Not that $p$ is true, or that $q$ is true. You simply rule out that both $p$ is true and $q$ is false.
But when I apply a method of mathematical induction, for example, I need to prove the hypothesis in order to show that the conclusion in true.
Wrong. Assuming is not proving. You only need to assume that the hypothesis is true.
You assume, for example, that $P(n)$ is true for some unspecified $n\in N$. Then prove that $P(n+1)$ would then also be true.
You could then generalize that, for all $n\in N$, you would have $P(n) \implies P(n+1).$
With $P(0)$ being true, you would then have, for all $n\in N, ~P(n)$.
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Right, assuming is not the same as proving, but why would you want to assume something if you want to prove nothing? You can treat mathematical induction as the statement of the form $p \implies q$, where $p$ is a compound statement and $q$ is the conclusion you want to prove. In other words, I need to prove $p$ to deduce $q$. – Vlad Mikheenko Nov 08 '24 at 09:04
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@VladMikheenko You want to prove the implication, not the antecedent. You start by assuming, or asking "what if $P(n)$ was true for some $n\in N$." Then you need to prove that $P(n+1)$ would then also have to be true. Then you can conclude that $P(n) \implies P(n+1)$ for ANY $n\in N$. – Dan Christensen Nov 08 '24 at 14:55
Adding to Karl's and Mauro's remarks.
Whatever mathematical statement that you're proving (whether it's of the form A ⟹ B with hypothesis A, or not a conditional), your assumptions include mathematical theorems.
When you're using mathematical induction to prove ∀n P(n), the hypotheses are actually the mathematical theorems that you're implicitly invoking, together with the base case P(0) and the induction step's conclusion ∀n∈N (P(n) ⟹ P(n+1)), which is derived from the induction step's own hypothesis For an arbitrary natural number k, P(k) is true, which is not being proven true within the induction step.
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