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I don't know how to prove that $p\Rightarrow q$ is equivalent to $\neg p\lor q$ ,here is the link p=>q . And I don't know how wolframalpha generate "Minimal forms" .

Can you prove $p\Rightarrow q \equiv \neg p\lor q$, and explain how to get "Minimal forms" ?

Thanks!

Asaf Karagila
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Frei Zhang
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3 Answers3

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By definition, $p\Rightarrow q$ is true if and only if the consequence $q$ is true, or the antecedent $p$ is false. You can see it in the truth table that defines the implication. That is, $p\Rightarrow q$ is true if and only if either $\neg p$ is true, or $q$ is true; i.e., if and only if $\neg p\lor q$ is true (what you write as $!p||q$).

Or you can simply look at the truth tables. The truth table of $\neg p\lor q$ is the same as the truth table of $p\Rightarrow q$: true if $p$ and $q$ are false; true if $p$ is false and $q$ is true; false if $p$ is true and $q$ is false; true if $p$ and $q$ are both true: $$\begin{array}{c|c||c} p & q & p\Rightarrow q\\ \hline 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 0 & 0\\ 1 & 1 & 1 \end{array}\qquad\qquad \begin{array}{c|c|c|c} p & q & \neg p & \neg p\lor q\\ \hline 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 1 \end{array}.$$ The final columns are identical, so the two formulas take the same truth values given the same truth inputs: that is, they are propositionally equivalent.

Arturo Magidin
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  • Thanks a lot! have a nice day:) – Frei Zhang Aug 18 '11 at 03:56
  • @ArturoMagidin The question is why would the logical implication φ ⇒ ψ, read and understood as "If φ, then ψ", have the same truth values as the material implication φ ⊃ ψ. – Speakpigeon Aug 12 '20 at 07:39
  • @Speakpigeon: That might be your question, but it was clealry not the question that was asked and answered nine years ago. Thanks for the downvote, though, based on the fact that a decade ago I did not anticipate your wishes and desires. – Arturo Magidin Aug 12 '20 at 14:02
  • @ArturoMagidin Where does the question say "material implication"? – Speakpigeon Aug 12 '20 at 14:07
  • @Speakpigeon: The fact that the answer was accepted and that the person asking the question did not make any clarification should speak for itself. I’m very sorry that I failed to read your mind prospectively back then. – Arturo Magidin Aug 12 '20 at 14:44
  • @ArturoMagidin You didn't have to read my mind. You only had to understand what the question meant from what it said. Where does the question says "material implication"? – Speakpigeon Aug 12 '20 at 15:16
  • @Speakpigeon: Again: the fact that I interpreted the question in the standard mathematical sense, and that the asker agreed with that interpretation should speak for itself. You’ve made your position clear: I was wrong to not read the question the way you think I should have, and I was wrong to answer the question to the asker’s satisfaction in a way that fails to satisfy you. Good for you. Consider your downvote your win, and you may take you ball and go home. Where does the question say I should answer your interpretation of it? Nowhere. You’ve emptily made your point. Move on. – Arturo Magidin Aug 12 '20 at 15:38
  • @Speakpigeon: The “material implication” is the standard meaning of both $\implies$ and $\rightarrow$ in propositional calculus. The question is tagged propositional calculus. it doesn’t take a leap to interpret the symbol in a standard way. So sorry you aren’t satisfied. – Arturo Magidin Aug 12 '20 at 15:44
  • @ArturoMagidin Wikipedia says: "A ⇒ B means if A is true then B is also true". – Speakpigeon Aug 12 '20 at 16:47
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Using truth tables is a simple way to prove it.

Quinn Culver
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I don't reckon that you need a Truth Table. What do you reckon of the intuitive explanation beneath?

From: Philip Johnson-Laird BA PhD Psychology (UCL), Stuart Professor of Psychology Emeritus at Princeton. (Author isn't a logician.) How We Reason (1st edn 2008). p. 108.

I changed the author's choice of first names, to ones that start with P and Q to fit the title. I symbolized the disjunctions in square brackets.

An exclusive disjunction, such as:

Either Pia helped or Quinn helped, but not both

is equivalent to the proposition:

Pia helped or the Quinn helped, and not both Pia helped and the Quinn helped.

Hence, exclusive disjunction also has a logical meaning.
In an analogous way we can define a logical meaning of “if”. The sentence:

If Pia didn’t help then Quinn did. [If ¬P, then Q.]

means:

Pia helped or Quinn did, or both. [P ∨ Q]

In its logical meaning, the conditional is compatible with three possibilities: Pia didn’t help and Quinn did [¬P ∧ Q] , Pia helped and Quinn didn’t [P ∧ ¬Q], Pia helped and Quinn helped [P ∧ Q]. The only possibility that the conditional rules out is that neither the Pia nor Quinn helped [¬P ∧ ¬Q]. The three possibilities that the conditional allows are the same as those for the inclusive disjunction.

  • You have only proved the trivial property (p → q) → (¬p ∨ q), which follows anyway from the definition of p → q as "If p, then q". To prove the equivalence (p → q) ⇔ (¬p ∨ q), you now need to prove (¬p ∨ q) → (p → q). – Speakpigeon Aug 12 '20 at 07:38