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Prove that $((p\lor q)\land(p\lor(\lnot q)))\rightarrow p$

Please could someone give me some feed back on this proof? Does it look correct?

= $\lnot ((p\lor q)\land(p\lor(\lnot q)))\lor p$

= $ (\lnot(p\lor q) \lor \lnot(p\lor(\lnot q)))\lor p$

= $((\lnot p\land \lnot q)\lor(\lnot p\land \lnot(\lnot q)))\lor p$

= $((\lnot p\land \lnot q)\lor(\lnot p\land q))\lor p$

= $(\lnot p(\lnot q \lor q)))\lor p$ <----- Is this step correct?

= $(\lnot p(T))\lor p$ $\equiv true $

Thanks a million

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    $\begingroup$ @amWhy has already pointed out the actual error, but I’ll note in addition that the fourth line duplicates the third. $\endgroup$ – Brian M. Scott Jan 17 '13 at 17:36
  • $\begingroup$ Unless you're required to stick with a particular proof formalism, a plain truth table proof might be shorter and simpler. $\endgroup$ – Henning Makholm Jan 17 '13 at 18:07
  • $\begingroup$ (Though not shorter and simpler than amWhy's answer) $\endgroup$ – Henning Makholm Jan 17 '13 at 18:15
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The step in question should be, using the distributive law: $$((\lnot p\land \lnot q)\lor(\lnot p\land q))\lor p$$ $$\equiv (\lnot p \land (\lnot q \lor q)) \lor p)$$

$$\equiv (\lnot p \land T) \lor p$$ $$\equiv \lnot p \lor p \equiv T$$

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    $\begingroup$ You're welcome: everything else looks good; just be sure to include your "justifications" for each step (DeMorgan's Law, etc.) in your final proof. $\endgroup$ – Namaste Jan 17 '13 at 17:44
  • $\begingroup$ Will do thanks.. $\endgroup$ – bosra Jan 17 '13 at 17:45

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