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" ?


  • $\begingroup$ Okay, => means implication, $\Rightarrow$. I take it that $||$ means "or", $\lor$? $\endgroup$ – Arturo Magidin Aug 18 '11 at 3:39
  • $\begingroup$ yes thanks I'm new here $\endgroup$ – ifree Aug 18 '11 at 3:42
  • $\begingroup$ No problem. Want me to edit it to more standard mathematical notation? $\endgroup$ – Arturo Magidin Aug 18 '11 at 3:45
  • $\begingroup$ yes thanks and how about the Minimal forms $\endgroup$ – ifree Aug 18 '11 at 3:49
  • $\begingroup$ There are algorithms for transforming propositional formulas into normal forms; see for example Wikipedia. Presumably, wolframalpha has programmed the algorithms in order to find the normal minimal forms. $\endgroup$ – Arturo Magidin Aug 18 '11 at 3:52

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.

  • $\begingroup$ Thanks a lot! have a nice day:) $\endgroup$ – ifree Aug 18 '11 at 3:56

Using truth tables is a simple way to prove it.


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.]


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.


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