Fallacy Equation

For the above proposition, I know it is a fallacy from the truth table when A is False and B is True.

How do I prove this fallacy using natural deduction, i.e for A = ꓕ and B = T, prove the above proposition as False


$((ꓕ \supset ꓔ) \supset (¬ꓕ \supset ¬ꓔ)) \supset ꓕ$ is true

  • $\begingroup$ I can't see the | |, dunno if that is a or b $\endgroup$ – asdf334 Sep 11 at 2:32
  • $\begingroup$ The proposition is-: (A => B) => (¬A => ¬B) When A is false and B is true, we need to deduce the above proposition as false Let me know if you still have any problem understanding $\endgroup$ – hjuk12 Sep 11 at 2:57

$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\def\imp{\supset}$

Indeed.   The proposition is contingent; it is false when $\neg A\wedge B$, and true otherwise.

So you must prove $\neg A\wedge B, (A\imp B)\imp(\neg A\imp\neg B)\vdash \bot$.

Here is a Fitch style ND skeleton.

$$\fitch{~~1.~\neg A\wedge B\\~~2.~(A\imp B)\imp(\neg A\imp \neg B)}{~~3.~\neg A\hspace{12ex}\wedge\mathsf E~1\quad\textsf{(Simplification)}\\~~4.~B\hspace{13.5ex}\wedge\mathsf E~1\quad\textsf{(Simplification)}\\\fitch{~~5.~}{~~6.~}\\~~7.~\\~~8.~\\~~9.~\neg B\\10.~\bot\hspace{14ex}\neg\mathsf E~4,9\quad\textsf{(Contradiction)}}$$


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