Prove (p & !q) -> ~(p -> q)
Truth Table p & !q
p | q | p & !q
F | F | F
F | T | F
T | F | T
T | T | F
Truth Table !(p -> q)
p | q | !(p -> q)
F | F | F
F | T | F
T | F | T
T | T | F
Truth Table p & !q
p | q | p & !q
F | F | F
F | T | F
T | F | T
T | T | F
Truth Table !(p -> q)
p | q | !(p -> q)
F | F | F
F | T | F
T | F | T
T | T | F