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In my homework I was given this question:

  1. Prove that $p \land\lnot p$ is a fallacy.

All I knew is that to prove a question like this you must have, two premises and a conclusion

(For example: Prove that $p ∧ ¬Q, P Ⱶ ¬p$ is a fallacy or valid. Premise 1 = $(p ∧ ¬Q)$, Premise 2= $(p)$, Conclusion is = $(¬p),$ but in this case we only have this $(p ∧ ¬p).$

How can you find if the proposition is valid or a fallacy or it is satisfiable or unsatisfiable. anyone with an idea? Many thanks to everyone to help me out.

Thomas Andrews
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17 George
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  • I'm not sure it is technically true that $p ∧ ¬Q, P Ⱶ ¬p$ is a "fallacy." It depends on the meaning of the word "fallacy," but I'd only use the word "fallacy" for a propositional calculus phrase, and "$p ∧ ¬Q, P Ⱶ ¬p$" is a statement about a logical deduction in propositional calculus. The equivalent propositional calculus phrase, $((p\land\lnot Q)\land P)\implies\lnot p,$ is a fallacy. – Thomas Andrews Nov 02 '22 at 15:29
  • In the end, a statement $S$ is a fallacy if and only if $\lnot S$ is always true. (Truism? I can't recall the term for the opposite of fallacy.) So you want to prove $\lnot(p\land\lnot p)$ is always true. – Thomas Andrews Nov 02 '22 at 15:36
  • Is $p\wedge \lnot p$ a reasoning? – A. P. Nov 02 '22 at 15:42

2 Answers2

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There are only two possible logic values for $p$: either it is false, which corresponds to $p = 0$ or it is true, which corresponds to $p = 1$. In the first case, one has that $p\wedge\neg p = 0\wedge 1 = 0$. On the second case, one has that $p\wedge\neg p = 1\wedge 0 = 0$. Since we have exhausted all the possibilities, it can be concluded that $p\wedge\neg p$ is a fallacy, and we are done.

Átila Correia
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The typical way to prove this is by contradiction.

If we know $A\land B$ we can deduce $A$ and $B.$

If $p\land \lnot p$ is true (the premise), we can deduce both $p$ and $\lnot p,$ so we get a contraction.

In some sense, $p\land \lnot p$ is the 'most basic' contradiction, in that any time you can prove a contradiction, you can rephrase it in this form in one propositional calculus statement. (Normally, a contradiction is two statements, $S$ and $\lnot S.$)

But how you write this proof heavily depends on what type of proofs you use. Each course introducing propositional logic uses slightly differing sets of axioms and proof approaches. They are all equivalent.

(Aside: There are non-equivalent cases of propositional calculus, but they are never taught in introductory classes.)

Thomas Andrews
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