prove that (R>P)&(~Q>~R) and R>(P&Q) are equivalent withouth the use of truth ta
ID: 1949134 • Letter: P
Question
prove that (R>P)&(~Q>~R) and R>(P&Q) are equivalent withouth the use of truth tablesExplanation / Answer
Going the other way, first assume (P ^ Q) > R, then assume, on consecutive lines, P and Q. Conjoin these to get P ^ Q, then apply >E to get R. R will thus rest on your initial assumption, (P ^ Q) > R, plus your two further assumptions, P and Q. Discharge the latter two assumptions in turn, so that you first have Q > R resting on (P ^ Q) > R and P, then P > (Q > R) resting just on (P ^ Q) > R. Apply >I one last time to get ((P ^ Q) > R) > (P > (Q > R)). You thus have (P > (Q > R)) > ((P ^ Q) > R) and ((P ^ Q) > R) > (P > (Q > R)), which together yield, by I, (P > (Q > R)) ((P ^ Q) > R). Generally speaking, if there aren't any negation signs, you shouldn't need to use ~I or ~E. One exception to this is the proof of the theoremhood of Peirce's law, i.e., ((P > Q) > P) > P: having assumed the antecedent - that is, (P > Q) > P - you can't derive the consequent, P, directly: you need to assume ~P, then derive P > Q, and then finally apply >E to get P. You'll thus have a contradiction, allowing you to deny ~PRelated Questions
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