- #1
gnome
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- 1
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)
[/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?
[/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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