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Thread: Indirect proof

  1. #1
    eps
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    Indirect proof

    Hi everyone,
    I really need help to prove this statement:
    (p→s) from( p→not q), (r→q )and (not r→s)

    Thanks
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  2. #2
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    Re: Indirect proof

    (r→q ) is equivalent to (not q→not r ). The latter is the contrapositive.
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  3. #3
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    Re: Indirect proof

    Quote Originally Posted by eps View Post
    Hi eve
    need help to prove this statement: (p→s) from( p→not q), (r→q )and (not r→s)
    Here are the premises: $I~(p\to\neg q)~,~II~(r\to q)~\&~III,~\neg(r\to s).$

    To prove by indirect proof you must assume that $A.~(p\wedge \neg s)$ is true.
    From A. & I. you can get $\neg q$. HOW?
    Then from II. you get $\neg r$
    Then from III. you get $ r$
    So how has it been proven that $p\to s~?$
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  4. #4
    eps
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    Re: Indirect proof

    Thank Plato, actually the third promise is (r→s) but you wrote (r→s).
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  5. #5
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    Re: Indirect proof

    Quote Originally Posted by eps View Post
    Thank Plato, actually the third promise is (r→s) but you wrote (r→s).
    But you said not r implies s. The usual order of operations would be (not r) implies s if that is what you meant.
    It still works. from $\neg r$ and $\neg r\to s$ you get $s$ modus pomens.
    Then from that and A. you get $s\wedge\neg s$ DONE>
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  6. #6
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    Re: Indirect proof

    Quote Originally Posted by eps View Post
    Hi everyone,
    I really need help to prove this statement:
    (p→s) from( p→not q), (r→q )and (not r→s)

    Thanks
    To continue from " p→not q" my next implication would have to start "not q". So I recognize that r→q is equivalent to it contrapositive, "not q→ not r". Replacing "r→ q" with "not q→ not r", I have the chain
    "p→ not q", "not q→ not r" and "not r→ s" from which I immediately deduce "p→ s".
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