Using only modus ponens or substitution. Prove:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Challenging logic problem

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