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Challenging logic problem

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  1. Dec 13, 2015 #1
    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.
     
  2. jcsd
  3. Dec 13, 2015 #2

    fresh_42

    Staff: Mentor

    If p then it follows from 2). (s≡p ; p≡q ; q≡r) But isn't 3) just p? I guess my understanding of 3) is f.
    May I conclude f → (p → f) by 1) and then f → p by 3)? If so I'm done by 2) and the substitutions mentioned.
     
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