- #1
JustNick
- 1
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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 point in the right direction is appreciated.
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 point in the right direction is appreciated.