How is this done without conditional proof?

Question

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):

1. $(F \implies E) \land (F \land E \implies R) $ Premise
2. $(F \implies E) $ 1.
3. $(F \land E \implies R)$ 1.
4. $F$ Assumption
5. $E$ 2.
6. $F \land E$ 4, 5.
7. $R$ 3.
8. $F \implies R$ 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.

Answer 1

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} } $

Answer 2

Here's an outline, just fill in the gaps.

$\begin{array}{|rll}1.&(F\to E)\wedge((F\wedge E)\to R)&\text{Premise}\\\hline 2.&F\to E&\text{And elimination 1} \\3.&(F\wedge E)\to R&\text{And elimination 1}\\4.&F\to (F\wedge F)&\text{Tautology (Idempotence of And)}\\\vdots&\vdots&\vdots\\n{-}1.&{(F\wedge F)\to R}&\text{Exportation }\ldots\\n.&F\to R&\text{Hypothetical Syllogism 4,n-1}\end{array}$

Answer 3

I don't have a proof for this, but I have a feeling there might be no such formal proof possible. One of Hurley's rules is Absorption, which allows you to derive $P \to (P \land Q)$ from $P \to Q$, and that is exactly what seems to be needed here. Or at least: with Absorption you can then go from $F \to E$ to $F \to (F \land E)$, and so with a simple Hypothetical Syllogism you can combine that with $(F \land E) \to R$ to get $F \to R$. But without Absorption .... you seem to be up the creek!