language, logic and discrete math

[jimmy5023]

can anyone prove

[(p->q) ^ (q -> r)] 0> (p -> r) <=> T?

Don't use the truth table, use the implication

[ne555]

I guess this won't help you so
¬ = not
+ = or
* = and, ^
p->q = ¬p+q

((p->q) * (q->r))->(p->r)

¬((p->q) * (q->r)) + (p->r)
¬(p->q) + ¬(q->r) + ¬p + r
¬(¬p+q) + ¬(¬q+r) + ¬p + r
p*¬q + q*¬r + ¬p + r
r + q + ¬p + ¬q
q+¬q + ...
1

r + ¬r*q = r+q
¬p + p*¬q = ¬p+¬q

[ultifinitus]

can anyone prove

dear friend, the real question is can you prove ... that.

[Zhuge]

Well said.