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I am given a premise and its resolution. But I cannot understand the steps. The premise is:

~(p|q => q|s)

and when I put that to http://intrologic.stanford.edu/applications/converter.html I get the result below:

{p,q}, {~q}, {~s}

I tried to get the same result by my hand but failed. So far I reached:

~(p | q => q | s)
(p & ~q & ~s) | (r & ~q & ~s)
(p|r) & (p|~q) & (p|~s) & ~q & ~s

But I stucked here. How can I get the final answer from here?

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You can replace (a => b) with (~a | b):

~(p|q => q|s)

~(~(p|q) | (q|s))

(p|q) & (~(q|s))

(p|q) & (~q) & (~s)

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  • $\begingroup$ Also note that for this expression to be true, q and s must be false. Because q is false, the value of p|q is actually equalt to p. So this expressoin can be true only for p=true, q=s=false. $\endgroup$
    – Saša
    Apr 30, 2018 at 10:49
  • $\begingroup$ I see! I should've stopped at (p|q) & (~q) & (~s) but I went on further and found myself in the endless maze such as (p & ~q & ~s) | (r & ~q & ~s). Thank you for the insight! $\endgroup$
    – JSong
    Apr 30, 2018 at 12:48

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