Hurley, "…Logic" 8th ed. section 7.5 introduces conditional proof. Some exercises are designed to show proofs that are much easier using conditional proof. For example 7.5 I (2):
- $(F \implies E) \land (F \land E \implies R) $ Premise
- $(F \implies E) $ 1.
- $(F \land E \implies R)$ 1.
- $F$ Assumption
- $E$ 2.
- $F \land E$ 4, 5.
- $R$ 3.
- $F \implies R$ 4, 7.
I would like to see a proof of this without conditional proof. The allowable rules are these: Modus ponens, Modus tollens, Hypothetical syllogism, Disjunctive syllogism, Constructive Dilemma, And-introduction and elimination (named differently), Or-introduction, DeMorgan's laws, Commutivity, Associativity, distribution, double negation, Transposition (contrapositive), Material implication, Exportation $ (P \land Q) \implies R \iff (P \implies (Q \implies R))$, and tautologies $p \iff p \land p$, and $p \iff p \lor p$.
Translating things using material implication and then reducing using DeMorgan's and distribution is not working out for me. Constructive dilemma could be used on $ \lnot E \implies \lnot F $ and $E \implies (\lnot F \lor R))$ if we could introduce $E \lor \lnot E$, but there is no rule allowing this.
Best Answer
I think a possible proof in the style of Hurley's book (without using Conditional Proof), could be:
$ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} $ $ \fitch{1.\,F \supset E\\ 2.\,(F \bullet E) \supset R \qquad \backslash F \supset R }{ 3.\,\lnot R \supset \lnot(F \bullet E) \qquad 2, \text{Trans}\\ 4.\,\lnot \lnot R \lor \lnot(F \bullet E)\qquad 3, \text{Impl}\\ 5.\,R \lor \lnot(F \bullet E)\qquad 4, \text{DN}\\ 6.\,R \lor (\lnot F \lor \lnot E)\qquad 5, \text{DM}\\ 7.\,(R \lor \lnot F) \lor \lnot E\qquad 6, \text{Assoc}\\ 8.\,\lnot E \lor (R \lor \lnot F)\qquad 7, \text{Com}\\ 9.\,\lnot E \lor (\lnot F \lor R)\qquad 8, \text{Com}\\ 10.\,E \supset (\lnot F \lor R)\qquad 9, \text{Impl}\\ 11.\,E \supset (F \supset R)\qquad 10, \text{Impl}\\ 12.\,F \supset (F \supset R)\qquad 1,11\,\text{HS}\\ 13.\,(F \bullet F) \supset R\qquad 12, \text{Exp}\\ 14.\,F \supset R \qquad 13, \text{Taut} } $