Question

Prove: (AvB)-->C /:. A-->C

Answer 1

pu: A v B

p2: A -> C

p3: B -> D

assume ~(C v D)~C & ~D (from 1, De Morgan's law)~C (from 2, conjunction elimination)~D (from 2, conjunction elimination)~A (from 3, p2, modus tollens)

Answer 2

Answer:

pu2-51 Pru)85#tttttttt