I want to prove [itex](A \supset B) \wedge (B \supset C) \wedge (D \supset \neg C) \wedge (A \vee D) \equiv (B \vee \neg C)(adsbygoogle = window.adsbygoogle || []).push({});

[/itex]

so I have to show that [itex]\neg ( ((A \supset B) \wedge (B \supset C) \wedge (D \supset \neg C) \wedge (A \vee D)) \supset (B \vee \neg C))[/itex]

is inconsistent, and I proceed as follows:

[tex]\begin{array}{ccccccccccc}\neg ( ((A \supset B) &\wedge &(B \supset C) &\wedge &(D \supset \neg C) &\wedge &(A \vee D)) &\supset (B \vee \neg C))\\

\neg ( \neg((A \supset B) &\wedge &(B \supset C) &\wedge &(D \supset \neg C) &\wedge &(A \vee D)) &\vee &(B \vee \neg C))\\

((A \supset B) &, &(B \supset C) &, &(D \supset \neg C) &, &(A \vee D)) &, &\neg (B \vee \neg C))\\

(\neg A \vee B) &, &(\neg B \vee C) &, &(\neg D \vee \neg C) &, &(A \vee D) &, &\neg B&, &C))\\

\text{contradiction}&, &\neg B &, &\neg D&, &A &, &\neg B &, &C

\end{array}[/tex]

so I end up with a contradiction showing that the original statement is correct.

Question: is there a "better", more formal way to present this proof?

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# Propositional logic proof

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