# Prove proposition is a fallacy

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

OR

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

• I can't see the | |, dunno if that is a or b – asdf334 Sep 11 at 2:32
• 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 – 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 $$\neg A\wedge B, (A\imp B)\imp(\neg A\imp\neg B)\vdash \bot$$.
$$\fitch{~~1.~\neg A\wedge B\\~~2.~(A\imp B)\imp(\neg A\imp \neg B)}{~~3.~\neg A\hspace{12ex}\wedge\mathsf E~1\quad\textsf{(Simplification)}\\~~4.~B\hspace{13.5ex}\wedge\mathsf E~1\quad\textsf{(Simplification)}\\\fitch{~~5.~}{~~6.~}\\~~7.~\\~~8.~\\~~9.~\neg B\\10.~\bot\hspace{14ex}\neg\mathsf E~4,9\quad\textsf{(Contradiction)}}$$