If Σ, Σ+∈ SENT, θ* ∈ CTERM\(ST(Σ) ∪ ST(Σ+)) and θ§∈ ATERM and if [θ*, θ§, Σ] = [θ*,
θ§
, Σ+], then Σ = Σ+.
Proof: The theorem is proved analogously to the negation-case in the proof of Theorem
1-20 by applying Theorem 1-20 and Theorem 1-12. ■
Theorem 1-22.Unique substitution bases (b) for terms
If θ, θ+∈ TERM, θ* ∈ CTERM\(ST(θ) ∪ ST(θ+)), ξ∈ VAR, β∈ PAR and [θ*, ξ, θ] = [θ*, β,
θ+
], then θ+ = [β, ξ, θ].
Proof: By induction on the complexity of θ. Suppose θ ∈ ATERM. Now, suppose θ+ ∈
TERM, θ* ∈ CTERM\(ST(θ) ∪ ST(θ+)), ξ∈ VAR, β∈ PAR and [θ*, ξ, θ] = [θ*, β, θ+].
Then we have θ ∈ CONST ∪ PAR ∪ VAR. Now, suppose θ∈ CONST. Then we have
[θ*, ξ, θ] = θ. Then we have θ = [θ*, β, θ+]. Because of θ* ∉ ST(θ) and Theorem 1-14-(i), we then have that θ = θ+ and because of θ ≠ξ we have θ+ = θ = [β, ξ, θ]. Now, suppose θ
∈ PAR. Then we have [θ*, ξ, θ] = θ. Then we have θ = [θ*, β, θ+]. Because of θ* ∉ ST(θ) and Theorem 1-14-(i), we then have again θ = θ+ and because of ξ≠θ: θ+ = θ = [β, ξ, θ].
Now, suppose θ∈ VAR. Suppose θ = ξ. Then we have [θ*, ξ, θ] = θ*. Then we have θ* =
[θ*, β, θ+]. Because of θ* ≠ θ+, we then have β∈ ST(θ+). Thus we have θ* ∈ ST([θ*, β, θ+
]). If θ+ ≠β, we would have, with θ* = [θ*, β, θ+], that θ* is a proper subterm of itself,
which contradicts Theorem 1-8. Therefore we have θ+ = β = [β, ξ, θ]. Now, suppose θ≠ξ.
Then we have θ = [θ*, ξ, θ]. Then we have θ = [θ*, β, θ+]. Because of θ* ∉ ST(θ) and Theorem 1-14-(i), we then have θ = θ+ and, because of θ≠ξ, we thus have θ+ = θ = [β, ξ, θ].
Now, suppose the statement holds for {θ0, …, θr-1} ⊆ TERM and suppose φ(θ0, …,
θr-1) ∈ FTERM. Now, suppose θ+ ∈ TERM, θ* ∈ TERM\(ST( φ(θ0, …, θr-1) ) ∪
ST(θ+)), ξ∈ VAR, β∈ PAR and [θ*, ξ, φ(θ0, …, θr-1) ] = [θ*, β, θ+]. Therefore [θ*, β,
θ+
] = φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) ∈ FTERM. Suppose for contradiction that θ+ ∈
ATERM. We have β≠ θ+
or β = θ+. Suppose β≠ θ+
. Then we have θ+ = [θ*, β, θ+] = φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) ∈ FTERM. Contradiction! Suppose β = θ+. Then we have
θ* = [θ*, β, θ+] = φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) . With Theorem 1-14-(i), it then follows
that for all i < r: [θ*, ξ, θi] = θi or there is an i < r such that θ* ∈ ST([θ*, ξ, θi]). If [θ*, ξ,
θr-1) and thus θ* ∈ ST( φ(θ0, … θr-1) ), which contradicts the hypothesis. If, on the other
hand, there was an i < r such that θ* ∈ ST([θ*, ξ, θi]), then θ* would be a proper subterm of φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) and therefore a proper subterm of itself, which contra-
dicts Theorem 1-8. Therefore θ+∉ ATERM, but θ+ ∈ FTERM. Therefore there are {θ'0,
…, θ'k-1} ⊆ TERM and φ' ∈ FUNC such that θ+ = φ'(θ'0, …, θ'k-1) . Thus we have
φ'([θ*, β, θ'0], …, [θ*, β, θ'k-1]) = [θ*, β, φ'(θ'0, …, θ'k-1) ] = [θ*, β, θ+] = φ([θ*, ξ, θ0],
…, [θ*, ξ, θr-1]) . With Theorem 1-11-(ii), it then follows that k = r and φ' = φ and [θ*, β,
θ'i] = [θ*, ξ, θi] for all i < r. With the I.H., it follows that θ'i = [β, ξ, θi] for all i < r. Thus we have θ+ = φ'(θ'0, …, θ'k-1) = φ([β, ξ, θ0], …, [β, ξ, θr-1]) = [β, ξ, φ(θ0, …, θr-1) ]. ■
Theorem 1-23.Unique substitution bases (b) for formulas
If Δ, Δ+∈ FORM, θ* ∈ TERM\(ST(Δ) ∪ ST(Δ+)), ξ∈ VAR, β∈ PAR and [θ*, ξ, Δ] = [θ*, β,
Δ+
], then Δ+ = [β, ξ, Δ].
Proof: Let Δ, Δ+∈ FORM, θ* ∈ CTERM\(ST(Δ) ∪ ST(Δ+)) and ξ∈ VAR, β∈ PAR and
[θ*, ξ, Δ] = [θ*, β, Δ+]. In the same way as we did in the inductive step of the preceding
proof for functional terms, one can show for all formulas that substitution bases (Δ and
Δ+
) belong to the same category and have the same main operator (predicate, connective or quantifier) as the respective substitution results ([θ*, ξ, Δ] and [θ*, β, Δ+]). The proof is
carried out by induction on the complexity of Δ. Suppose Δ = Φ(θ0, … θr-1) ∈
AFORM. Then we also have [θ*, ξ, Δ] = Φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) ∈ AFORM and
there are {θ'0, …, θ'r-1} ⊆ TERM with Φ(θ'0, …, θ'r-1) = Δ+. Therefore we also have
Φ([θ*, ξ, θ0], …, [θ*, ξ, θr-1]) = [θ*, ξ, Δ] = [θ*, β, Δ+] = [θ*, β, Φ(θ'0, …, θ'r-1) ] =
Φ([θ*, β, θ'0], …, [θ*, β, θ'r-1]) ∈ AFORM. With Theorem 1-11-(iv), it then follows
that [θ*, ξ, θi] = [θ*, β, θ'i] for all i < r. With Theorem 1-22, it follows that θ'i = [β, ξ, θi] for all i < r. Thus we then have Δ+ = Φ(θ'0, … θ'r-1) = Φ([β, ξ, θ0], …, [β, ξ, θr-1]) =
[β, ξ, Φ(θ0, …, θr-1) ] = [β, ξ, Δ].
Now, suppose the statement holds for Δ0, Δ1 ∈ FORM and let Δ = ¬Δ0 ∈
CONFORM. Then we also have [θ*, ξ, Δ] = ¬[θ*, ξ, Δ0] ∈ CONFORM and there is
Δ'0∈ FORM with ¬Δ'0 = Δ+. Therefore we also have ¬[θ*, ξ, Δ0] = [θ*, β, Δ+] = [θ*,
β, ¬Δ'0 ] = ¬[θ*, β, Δ'0] ∈ CONFORM. With Theorem 1-11-(v), it then follows that
[θ*, ξ, Δ0] = [θ*, β, Δ'0]. With the I.H., it follows that Δ'0 = [β, ξ, Δ0] and thus that Δ+ =
¬Δ'0 = ¬[β, ξ, Δ0] = [β, ξ, ¬Δ0 ] = [β, ξ, Δ]. Suppose Δ = (Δ0 ψ Δ1) ∈
and there are Δ'0, Δ'1∈ FORM with (Δ'0ψΔ'1) = Δ+. Therefore we also have ([θ*, ξ,
Δ0] ψ [θ*, ξ, Δ1]) = [θ*, β, Δ+] = [θ*, β, (Δ'0ψΔ'1) ] = ([θ*, β, Δ'0] ψ [θ*, β, Δ'1]) ∈
CONFORM. With Theorem 1-11-(vi), it then follows that [θ*, ξ, Δ0] = [θ*, β, Δ'0] and
[θ*, ξ, Δ1] = [θ*, β, Δ'1]. With the I.H., it follows that Δ'0 = [β, ξ, Δ0] and Δ'1 = [β, ξ, Δ1]
and thus we have Δ+ = (Δ'0ψΔ'1) = ([β, ξ, Δ0] ψ [β, ξ, Δ1]) = [β, ξ, (Δ0ψΔ1) ] = [β,
ξ, Δ].
Suppose Δ = Πξ'Δ0 ∈ QFORM. Suppose ξ' = ξ. Then we have Δ = Πξ'Δ0 = [θ*, ξ,
Πξ'Δ0 ] = [θ*, ξ, Δ] = [θ*, β, Δ+]. With Theorem 1-14-(ii), we then have θ* ∈ ST([θ*,
β, Δ+]) = ST(Δ) or [θ*, β, Δ+] = Δ+. This first case is excluded by the hypothesis. In the second case, we have that Δ+ = Πξ'Δ0 = [β, ξ, Πξ'Δ0 ] = [β, ξ, Δ]. Suppose ξ' ≠ ξ.
Then we have [θ*, ξ, Δ] = Πξ'[θ*, ξ, Δ0] ∈ QFORM and there is Δ'0∈ FORM with
Πξ'Δ'0 = Δ+. Therefore we also have Πξ'[θ*, ξ, Δ0] = [θ*, β, Δ+] = [θ*, β, Πξ'Δ'0 ] =
Πξ'[θ*, β, Δ'0] ∈ QFORM. With Theorem 1-11-(vii), it then follows that [θ*, ξ, Δ0] =
[θ*, β, Δ'0]. With the I.H., it follows that Δ'0 = [β, ξ, Δ0] and thus Δ+ = Πξ'Δ'0 = Πξ'[β,
ξ, Δ0] = [β, ξ, Πξ'Δ0 ] = [β, ξ, Δ]. ■
Theorem 1-24.Cancellation of parameters in substitution results
If θ∈ TERM, Δ∈ FORM, Σ∈ SENT, θ* ∈ CTERM, β∈ PAR\(ST(θ) ∪ ST(Δ) ∪ ST(Σ)) and
θ+∈
ATERM, then:
(i) [θ*, θ+, θ] = [θ*, β, [β, θ+, θ]],
(ii) [θ*, θ+, Δ] = [θ*, β, [β, θ+, Δ]], and
(iii) [θ*, θ+, Σ] = [θ*, β, [β, θ+, Σ]].
Proof: Let θ∈ TERM, Δ∈ FORM, Σ∈ SENT, θ* ∈ CTERM, β∈ PAR\(ST(θ) ∪ ST(Δ)
∪ ST(Σ)) and θ+∈ ATERM. Ad (i): The proof is carried out by induction on the complex-
ity of θ. Suppose θ ∈ ATERM. Then we have θ = θ+ or θ ≠ θ+
. First, suppose θ = θ+. Then we have [β, θ+, θ] = β and [θ*, θ+, θ] = θ*. Then we have [θ*, θ+, θ] = θ* = [θ*, β, β] = [θ*, β, [β, θ+, θ]]. Now, suppose θ≠θ+
. Then we have [β, θ+, θ] = θ and [θ*, θ+, θ] = θ. Because of β∉ ST(θ), we have β≠θ and thus θ = [θ*, β, θ]. Therefore we have [θ*, θ+, θ] = θ = [θ*, β, θ] = [θ*, β, [β, θ+, θ]].
Now, suppose the statement holds for {θ0, …, θr-1} ⊆ TERM and suppose θ = φ(θ0, …
θr-1) ∈ FTERM. Because of β∉ ST(θ), we also have that β∉ ST(θi) for all i < r. With the I.H., it then holds that [θ*, θ+, θi] = [θ*, β, [β, θ+, θi]] for all i < r. Then we have [θ*,
θ+
, φ(θ0, … θr-1) ] = φ([θ*, θ+, θ0], …, [θ*, θ+, θr-1]) = φ([θ*, β, [β, θ+, θ0]], …, [θ*, β,
[β, θ+, θr-1]]) = [θ*, β, φ([β, θ+, θ0], …, [β, θ+, θr-1]) ] = [θ*, β, [β, θ+, φ(θ0, … θr-1) ]].
Ad (ii): The proof is carried out by induction on the complexity of Δ. Suppose Δ =
Φ(θ0, … θr-1) ∈ AFORM. Then we have β∉ ST(θi) for all i < r and [θ*, θ+, Δ] = [θ*,
θ+
, Φ(θ0, … θr-1) ] = Φ([θ*, θ+, θ0], … [θ*, θ+, θr-1]) . With (i), it holds that [θ*, θ+, θi] = [θ*, β, [β, θ+, θi]] for all i < r. Therefore we have [θ*, θ+, Δ] = Φ([θ*, β, [β, θ+, θ0]], …,
[θ*, β, [β, θ+, θr-1]]) = [θ*, β, Φ([β, θ+, θ0], …, [β, θ+, θr-1]) = [θ*, β, [β, θ+, Φ(θ0, …
θr-1) ]] = [θ*, β, [β, θ+, Δ]].
Now, suppose the statement holds for Δ0, Δ1 ∈ FORM. First, let Δ = ¬Δ0 ∈
CONFORM. Then we have β ∉ ST(Δ0) and [θ*, θ+, Δ] = [θ*, θ+, ¬Δ0 ] = ¬[θ*, θ+,
Δ0] . With the I.H., it holds that [θ*, θ+, Δ0] = [θ*, β, [β, θ+, Δ0]]. Therefore [θ*, θ+, Δ] =
¬[θ*, β, [β, θ+, Δ0]] = [θ*, β, [β, θ+, ¬Δ0 ]] = [θ*, β, [β, θ+, Δ]]. Suppose Δ = (Δ0ψ
Δ1) ∈ CONFORM. This case is proved analogously to the negation-case.
Suppose Δ = ΠξΔ0 ∈ QFORM. Suppose ξ = θ+. Then we have [θ*, θ+, Δ] = [θ*, θ+,
ΠξΔ0 ] = ΠξΔ0 = [β, θ+, ΠξΔ0 ] = [β, θ+, Δ]. Then we have β ∉ ST([β, θ+, Δ]) =
ST(Δ). Therefore [θ*, θ+, Δ] = [β, θ+, Δ] = [θ*, β, [β, θ+, Δ]]. Suppose ξ≠θ+
. This case is proved analogously to the negation-case.
Ad (iii): This case is proved analogously to the negation-case. ■