0

Does anyone know how to do this without the use of addition rules? We have not covered that in class, and all the info I can find online suggests that as a solution. Thanks![]1

  • I provided a proof using a Fitch-style proof checker in this answer: philosophy.stackexchange.com/a/67381/29944 It uses different symbols, but otherwise it is the same. substitute R with A, substitute S with B, substitute P with C and substitute Q with D. – Frank Hubeny Oct 7 '19 at 8:59
0

You do not need a proof by contradiction. It is purely a proof by cases.

Just use disjunction introduction to achieve the required derivation under the assumed cases. Then use disjunction elimination to discharge the assumptions.

When you derive P you may derive PvQ by disjunction elimination

Likewise when you derive Q you may derive PvQ.

When you can derive PvQ under assumption of R, and can derive PvQ under assumption of S, then you may derive PvQ from RvS, via disjunction elimination.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.