If you did that at least in the way the author understands, the theory would no longer come as sound.
Note that the definition of the set Γ doesn't say anything about its members as axioms or coming as derivable from the axioms. The first clause of definition 2.2.3 says "a formula A$_i$∈Γ" What is Γ? The text says "the set of formulae Γ". The author doesn't clarify, but it does work out that Γ can be any finite subset of the set of all propositional formulas. So, yes you can apply uniform substitution to an element, or even some elements of Γ provided that those element(s) you apply uniform substitution to qualify as either an axiom, a theorem, or the negation of a theorem. But, you can't apply uniform substitution to all elements of Γ without mangling things. More specifically, uniform substitution in terms of deductions can't get applied to semantic contingencies.
Examples of how you might apply uniform substitution to elements of Γ. Suppose Γ contains only C-Kpq-q and K.Kpq.r as formulas (the "-"s and "."s are informal punctuation marks to hopefully make these formulas easier to read... K stands for logical conjunction, C for the material conditional here and I use Polish notation). Then we can write a deduction like this:
- K.Kpq.r -----hyothesis 1
- C-Kpq-q -----hypothesis 2
- C-K.Kpq.r-r -----2 p/Kpq, q/r 3
- r -----1, 3 conditional-out 4
- C-C.Kpq.q-r -----2-4 conditional-in 5
$\vdash$C-KKpqr-C.CKpqq.r -----1-5 conditional introduction 6
Suppose we have C-p-C.Cpq.q and C-Cpq-C.Cqr.Cpr, and C.CCpqq.r as hypotheses. Then we can write a deduction like this:
- C-p-C.Cpq.q -----hypothesis 1
- C-Cpq-C.Cqr.Cpr -----hypothesis 2
- C.CCpqq.r -----hypothesis 3
- C-C p CCpqq-C.C CCpqq r.Cpr -----2 q/CCpqq 4
- C.C CCpqq r.Cpr -----1, 4 conditional-out 5
- Cpr -----3, 5 conditional-out, 6
- C.CCCpqqr.Cpr -----3-6 conditional-in 7
- C.CCpqCCqrCpr.CCCCpqqrCpr -----2-7 conditional-in 8
$\vdash$C-CpCCpqq-C.CCpqCCqrCpr.C:CCCpqqr:Cpr -----1-8 conditional-in 9
Suppose we have K-p-Np (the negation of a theorem), C-Kpq-p (a theorem), and q (a contingent proposition) as hypotheses. Then we can write:
- p -------hypothesis 1
- C-Kpq-p ------ hypothesis 2
- K-p-Np -------hypothesis 3
- K-Cpq-NCpq ------- 3, p/Cpq 4
- C-KCpqNCpq-Cpq ----- 2, p/Cpq, q/NCpq 5
- Cpq ------- 2, 3 conditional-out 6
- C-KpNp-Cpq ------3-6 conditional-in 7
- C-CKpqp-C.KpNp.Cpq ------2-7 conditional-in 8
$\vdash$C-p-C.CKpqp.C:KpNp:Cpq ------1-8 conditional-in 9
Now suppose you were to apply uniform substitution to any possible permissible element of Γ. Then, it would become possible to write deductions like the following:
1 C p q |hypothesis
2 CaCba |axiom
3 C CaCba q |1 p/CaCba, q/q where "x/y" indicates x has gotten substituted by y
4 q |2, 3 modus ponens
Since that system has the deduction meta-theorem (or equivalently conditional introduction), it follows that CCpqq would become a theorem in propositional calculus. But, if p is false and q is false, then CCpqq=CC000=C10=0. Consequently, such a theory would no longer come as sound. Also, you would have inference rules whereby you could "prove" the consequent of a conditional from the conditional itself.
Added: You can apply uniform substitution to contradictions given the deduction metatheorem, and not mangle things since contradictions are false for all interpretations. Contradictions always take on a designated value. Thus, any substitution instance of a contradiction is false also. So, when you discharge any introduced contradictions by conditional introduction, you'll end up with a tautology of classical propositional logic, and thus soundness holds.
Added: There does exist something of another exception, if the author were to introduce propositional logic with quantifiers. St. Jaskowski's original paper The Rules of Supposition in Formal Logic on what we call "natural deduction" has a substitution rule when you have what we call a universal quantifier under special conditions.
Thank you, Ominvium, for the right answer. Where I made a major mistake is not putting the clues (who said what) into sentences. I only considered each person's statement as a whole, and created sentences that merely captured whether each person was lying or not.
Here is the complete solution (version 1), described by Ominvium. Since the original question is from the Propositional Resolution (PR) section of my notes, the solution below is in that form.
(comments very much appreciated)
\begin{array}{lll}
A & = & \text{Adams did it} \\
B & = & \text{Brown did it} \\
C & = & \text{Clark did it} \\
p & = & \text{Brown knew victim} \\
q & = & \text{Brown was in town } \\
r & = & \text{Clark was in town} \\
t & = & \text{Adams was in town} \\
S_A & = & (\lnot A \land p) \\
S_B & = & (\lnot B \land \lnot p \land \lnot q) \\
S_C & = & (\lnot C \land q \land t) \\
\end{array}
Quick recap on Propositional Resolution (PR)
PR depends on clausal form :
\begin{array}{ll}
(p \lor q) = & \left\{p,q\right\} \\
(p \land q) = & \left\{p\right\} \\
& \left\{q\right\} \\
\end{array}
Key thing to note for this question is that following the PR procedure, if we get an empty clause {}, the sentences contain a contradiction. For example:
\begin{array}{ll}
(p \land \lnot p) = & \left\{ p \right\} \\
& \left\{ \lnot p \right\} \\
& \text{--------} \\
& \left\{\right\}
\end{array}
So in my approach, I took the three scenarios: 1) Adams is lying, 2) Brown is lying, or 3) Clark is lying, and worked through each to see which results in an empty clause.
We have 3 version that we need to check:
1) Adams is lying and the others are telling the truth:
\begin{array}{lll}
\lnot S_A & = \lnot ( \lnot A \land p ) & \qquad\qquad \\
& = ( A \lor \lnot p ) \\
\end{array}
\begin{array}{lll}
(S_B \land S_C ) \land \lnot S_A \\
1) \lnot B \\
2) \lnot p \\
3) \lnot q \\
4) \lnot C \\
5) q \\
6) t \\
7) \left\{A, \lnot p\right\} \\
\text{--------} \\
8) \left\{\right\}& \mbox{(3,5) Contradiction} \\
\end{array}
2) Brown is lying and the others are telling the truth:
\begin{array}{lll}
S_B & = \lnot ( \lnot B \land \lnot p \land \lnot q) & \qquad\qquad \\
& = \lnot ( B \lor p \lor q) & \\
\end{array}
\begin{array}{ll}
(S_A \land S_C ) \land \lnot S_B \\
1) \lnot A \\
2) p \\
3) C \\
4) q \\
5) t \\
6) \left\{B, p, q \right\} \\
\text{--------} \\
7) \mbox{No Contradiction} \\
\end{array}
3) Clark is lying and the others are telling the truth:
\begin{array}{lll}
\lnot S_C & = \lnot ( \lnot C \land q \land t ) & \qquad\qquad \\
& = ( C \lor \lnot q \lor \lnot t ) \\
\end{array}
\begin{array}{lll}
(S_A \land S_B ) \land \lnot S_C \\
1) \lnot A \\
2) p \\
3) \lnot B \\
4) \lnot p \\
5) \lnot q \\
6) \left\{C, \lnot p, \lnot q \right\} \\
\text{--------} \\
7) \left\{\right\} & \mbox{(2,4) Contradiction} \\
\end{array}
Since (1) and (3) have contradictions, only (2) can be true.
$\therefore $ Brown is lying, and is the murderer.
Best Answer
$\{A1, A2, A3''\}$ cannot establish $A3$. Just consider when $\lnot$ is interpreted as the identity.
$\{A1, A2, A3'\}$ can establish $A3$ because it is classically complete. But the actual derivation can be long and tedious. For reference Bram's answer to this question: help with some Hilbert style proofs in a propositional logic axiom system.