Recent content by chisser98

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    Undergrad P=>q, q=>r, then p=>r (proof assumes p why?)

    Thanks for the reply Hurkyl. Ok - since S is not a consequence of S=>T, we can say it's true? I'm saying this because the implication introduction scheme is: M => (N => M) in this example, we've set M = (q=>r) and N = p, so we get: (q=>r) => (p => (q=>r)) I guess I'm just not...
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    Undergrad P=>q, q=>r, then p=>r (proof assumes p why?)

    hi guys, I'm struggling to learn logic and I'm stuck on what I'm sure is an easy answer. Here's the question: whenever p is true, q is true. Whenever q is true, r is true. Prove that, whenever p is true, r is true. Here's part of the answer: p=>q Premise q=>r Premise...