Recent content by JustNick

Using only modus ponens or substitution. Prove: q -> r -> [ [ p -> q ] -> [ p -> r ] ] using the three axioms: 1) p -> [ q -> p ] 2) s -> [ p-> q ] -> [ [s -> p] -> [ s -> q ] ] 3) p -> f -> f -> p where the symbol f is "false." I am having the hardest time trying to solve this proof, any...