I can't figure out how to prove that formally. Please, help!!

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2 Answers 2


In Fitch, how does one prove “P” from the premise “(¬P ∨ Q)→P”?

One assumes not-P and uses a Reduction To Absurdity proof.

|_ (~P v Q) -> P   Premise
|  |_ ~P           Assumption
|  |  :            :
|  |  :            :
|  |  :            :
|  ~~P             Negation Introduction
|  P               Double Negation Elimination

Here is a way to prove this using the rules in Klement's Fitch-style proof checker. The rules are described in forallx. Both are available in the links below and would make good supplementary material to whatever text you are using.

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This proof uses the law of the excluded middle (LEM). To use this I take a simple statement and its negation and from both try to derive the same result. If I get the same result than I can invoke the law of the excluded middle. Here I chose "P" and "¬P", because one of these, "P", is the goal itself.

For "P" I need do nothing but add the subproof with assumption "P". For "¬P" I use disjunction introduction to get line 4 and then conditional elimination on line 5 (modus ponens) to get "P". I reached the goal, "P", in both cases and so I can discharge the two assumptions on line 2 and 3 and reach the end of the proof.


Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

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