If θ ∈ TERM, Δ ∈ FORM, Σ ∈ SENT, ∈ SEQ, k ∈ N\{0} and {β*0, …, β*k} ⊆
PAR\(ST(θ) ∪ ST(Δ) ∪ ST(Σ) ∪ STSEQ( )) and {β0, …, βk} ⊆ PAR\{β*0, …, β*k}, where
β*i≠β*j and βi≠βj for all i, j < k+1 with i≠j, then:
(i) [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ]] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, θ],
(ii) [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Δ]] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, Δ],
(iii) [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Σ]] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, Σ], and
(iv) [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, ]] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, ].
Proof: Suppose θ ∈ TERM, Δ∈ FORM, Σ∈ SENT, ∈ SEQ, k∈ N\{0} and {β*0, …,
β*k} ⊆ PAR\(ST(θ) ∪ ST(Δ)) and {β0, …, βk} ⊆ PAR\{β*0, …, β*k}, where β*i ≠ β*j and βi≠ βj for all i, j < k+1 with i≠j. Ad (i): The proof is carried out by induction on the
complexity of θ. Suppose θ ∈ ATERM. Then we have θ ∈ CONST ∪ PAR ∪ VAR.
Now, suppose θ ∈ CONST ∪ VAR ∪ (PAR\{β0, …, βk}). Then we have θ = [〈β*0, …,
β*k-1〉, 〈β0, …, βk-1〉, θ] and we have θ = [〈β*0, …, β*k〉, 〈β0, …, βk〉, θ] and thus [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ]] = [β*k, βk, θ] = θ = [〈β*0, …, β*k〉, 〈β0, …, βk〉, θ].
Now, suppose θ ∈ {β0, …, βk}. Then we have θ = βi for an i < k+1. According to the hypothesis, we then have that for all j < k+1 with j ≠i it holds that θ≠ βj. Thus we have [〈β*0, …, β*k〉, 〈β0, …, βk〉, θ] = β*i. Now, suppose i < k. Then we have [〈β*0, …, β*k-1〉,
〈β0, …, βk-1〉, θ] = β*i and thus [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ]] = [β*k, βk, β*i]. By hypothesis, we have that βk≠β*i and thus that [β*k, βk, β*i] = β*i. Now, suppose i = k. Then we have [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ] = θ = βk and hence [β*k, βk, [〈β*0, …,
β*k-1〉, 〈β0, …, βk-1〉, θ]] = [β*k, βk, βk] = β*k = β*i.
Now, suppose the statement holds for {θ0, …, θr-1} ⊆ TERM and suppose θ = φ(θ0,
…, θr-1) ∈ FTERM. Then we have [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ]] = [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, φ(θ0, …, θr-1) ]] = φ([β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …,
βk-1〉, θ0]], …, [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θr-1]]) . With the I.H., it holds that
[β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θi]] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, θi] for all i < r. Therefore we have [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, θ]] = φ([〈β*0, …, β*k〉, 〈β0, …,
βk〉, θ0], …, [〈β*0, …, β*k〉, 〈β0, …, βk〉, θr-1]) = [〈β*0, …, β*k〉, 〈β0, …, βk〉, φ(θ0, …,
θr-1) ] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, θ].
Ad (ii): The proof is carried out by induction on the complexity of Δ. Suppose Δ =
Φ(θ0, … θr-1) ∈ AFORM. This case is proved analogously to the FTERM-case by ap-
Now, suppose the statement holds for Δ0, Δ1 ∈ FORM and suppose Δ = ¬Δ0 ∈
CONFORM. Then we have [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Δ]] = [β*k, βk, [〈β*0,
…, β*k-1〉, 〈β0, …, βk-1〉, ¬Δ0 ]] = ¬[β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Δ0]] . With
the I.H., it holds that [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Δ0]] = [〈β*0, …, β*k〉, 〈β0, …,
βk〉, Δ0]. Therefore we have [β*k, βk, [〈β*0, …, β*k-1〉, 〈β0, …, βk-1〉, Δ]] = ¬[〈β*0, …, β*k〉,
〈β0, …, βk〉, Δ0] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, ¬Δ0 ] = [〈β*0, …, β*k〉, 〈β0, …, βk〉, Δ].
Suppose Δ = (Δ0ψΔ1) ∈ CONFORM. This case is proved analogously to the negation-
case. Suppose Δ = ΠξΔ0 ∈ QFORM. This case is also proved analogously to the nega-
tion-case.
Ad (iii) and (iv): (iii) follows analogously to the negation-case by applying (ii), and (iv)
follows analogously to the FTERM-case by applying (iii). ■
Note: For sets of formulas, a theorem that is analogous to Theorem 1-27 can be proved.
Theorem 1-28.Multiple substitution of closed terms for pairwise different variables in terms
and formulas (a)
If k∈N\{0}, {θ*0, …, θ*k} ⊆ CTERM and {ξ0, …, ξk} ⊆ VAR, where ξi≠ξj for all i, j <
k+1 with i≠j, then:
(i) If θ∈ TERM, then
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θ], and
(ii) If Δ∈ FORM, then
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ].
Proof: Let k∈ N\{0}, {θ*0, …, θ*k} ⊆ CTERM and {ξ0, …, ξk} ⊆ VAR, where ξi ≠ξj
for all i, j < k+1 with i≠j. Ad (i): Suppose θ∈ TERM. The proof is carried out by induc-
tion on the complexity of θ. Suppose θ∈ ATERM. Suppose ξi ≠ θ for all i < k+1. Then
we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [θ*k, ξk, θ] = θ = [〈θ*0, …, θ*k〉, 〈ξ0,
…, ξk〉, θ]. Suppose ξi = θ for an i < k. Then we have ξj≠θ for all i < j < k+1. Then we have [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θ] = [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ] = [〈θ*0, …, θ*i〉, 〈ξ0,
…, ξi〉, θ] = θ*i∈ CTERM. Therefore [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [θ*k, ξk,
θ*i] = θ*i = [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θ]. Suppose ξk =
θ. Then we have ξi≠θ for all i < k and [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ] = θ. Therefore
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [θ*k, ξk, θ] = θ*k = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉,
Now, suppose the statement holds for {θ0, …, θr-1} ⊆ TERM and suppose θ = φ(θ0,
…, θr-1) ∈ FTERM. Then we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, φ(θ0, …, θr-1) ]] = φ([θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉,
θ0]], …, [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θr-1]]) . With the I.H., it holds that [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θi]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θi] for all i < r. Therefore we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = φ([〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θ0], …,
[〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θr-1]) = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, φ(θ0, …, θr-1) ] = [〈θ*0,
…, θ*k〉, 〈ξ0, …, ξk〉, θ].
Ad (ii): Suppose Δ∈ FORM. The proof is carried out by induction on the complexity of
Δ. Suppose Δ = Φ(θ0, … θr-1) ∈ AFORM. This case is proved analogously to the
FTERM-case by applying (i).
Now, suppose the theorem holds for Δ0, Δ1 ∈ FORM. Suppose Δ = ¬Δ0 ∈
CONFORM. Then we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = [θ*k, ξk, [〈θ*0, …,
θ*k-1〉, 〈ξ0, …, ξk-1〉, ¬Δ0 ]] = ¬[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ0]] . With the
I.H., it holds that [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ0]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉,
Δ0]. Therefore [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = ¬[〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉,
Δ0] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, ¬Δ0 ] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ]. Suppose Δ =
(Δ0ψΔ1) ∈ CONFORM. This case is proved analogously to the negation-case.
Suppose Δ = ΠζΔ0 ∈ QFORM. Suppose ξi = ζ for one i < k. Then we have ξj≠ ζ for all j < k+1 with i≠j. Then we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, ΠζΔ0 ]] = [θ*k, ξk, Πζ[〈θ*0, …, θ*i-1, θ*i+1, …, θ*k-1〉, 〈ξ0,
…, ξi-1, ξi+1, …, ξk-1〉, Δ0] ] = Πζ[θ*k, ξk, [〈θ*0, …, θ*i-1, θ*i+1, …, θ*k-1〉, 〈ξ0, …, ξi-1, ξi+1,
…, ξk-1〉, Δ0]] . With the I.H., it holds that [θ*k, ξk, [〈θ*0, …, θ*i-1, θ*i+1, …, θ*k-1〉, 〈ξ0, …,
ξi-1, ξi+1, …, ξk-1〉, Δ0]] = [〈θ*0, …, θ*i-1, θ*i+1, …, θ*k〉, 〈ξ0, …, ξi-1, ξi+1, …, ξk〉, Δ0]. There-
fore we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = Πζ[〈θ*0, …, θ*i-1, θ*i+1, …,
θ*k〉, 〈ξ0, …, ξi-1, ξi+1, …, ξk〉, Δ0] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, ΠζΔ0 ] = [〈θ*0, …, θ*k〉,
〈ξ0, …, ξk〉, Δ]. Suppose ξk = ζ. Then we have ξi ≠ ζ for all i < k and [θ*k, ξk, [〈θ*0, …,
θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, ΠζΔ0 ]] = [θ*k, ξk,
Πζ[〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ0] ] = Πζ[〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ0] = [〈θ*0,
…, θ*k〉, 〈ξ0, …, ξk〉, ΠζΔ0 ] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ].
Suppose ξi≠ζ for all i < k+1. Then we have [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] =
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, ΠζΔ0 ]] = [θ*k, ξk, Πζ[〈θ*0, …, θ*k-1〉, 〈ξ0, …,
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ0]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ0]. Therefore
[θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, Δ]] = Πζ[〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ0] = [〈θ*0,
…, θ*k〉, 〈ξ0, …, ξk〉, ΠζΔ0 ] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ]. ■
Theorem 1-29.Multiple substitution of closed terms for pairwise different variables in terms
and formulas (b)
If k∈N\{0}, {θ*0, …, θ*k} ⊆ CTERM and {ξ0, …, ξk} ⊆ VAR, where ξi≠ξj for all i, j <
k+1 with i≠j, then:
(i) If θ∈ TERM, then
[〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, [θ*k, ξk, θ]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, θ], and
(ii) If Δ∈ FORM, then
[〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, [θ*k, ξk, Δ]] = [〈θ*0, …, θ*k〉, 〈ξ0, …, ξk〉, Δ].
Proof: Suppose k∈ N\{0}, {θ*0, …, θ*k} ⊆ CTERM and {ξ0, …, ξk} ⊆ VAR, where ξi
≠ ξj for all i, j < k+1 with i≠ j. Ad (i): Suppose θ ∈ TERM. The proof is carried out by
induction on k. Suppose k = 1. With Theorem 1-25-(i) and Theorem 1-28-(i), we then
have [θ*0, ξ0, [θ*1, ξ1, θ]] = [θ*1, ξ1, [θ*0, ξ0, θ]] = [〈θ*0, θ*1〉, 〈ξ0, ξ1〉, θ]. Now, suppose 1
< k. Applying the I.H., Theorem 1-25-(i), the I.H., Theorem 1-28-(i), the I.H. and
Theorem 1-28-(i) (in this order) yields [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, [θ*k, ξk, θ]] = [〈θ*0,
…, θ*k-2〉, 〈ξ0, …, ξk-2〉, [θ*k-1, ξk-1, [θ*k, ξk, θ]]] = [〈θ*0, …, θ*k-2〉, 〈ξ0, …, ξk-2〉, [θ*k, ξk, [θ*k-1, ξk-1, θ]]] = [〈θ*0, …, θ*k-2, θ*k〉, 〈ξ0, …, ξk-2, ξk〉, [θ*k-1, ξk-1, θ]] = [θ*k, ξk, [〈θ*0, …,
θ*k-2〉, 〈ξ0, …, ξk-2〉, [θ*k-1, ξk-1, θ]]] = [θ*k, ξk, [〈θ*0, …, θ*k-1〉, 〈ξ0, …, ξk-1〉, θ]] = [〈θ*0, …,
θ*k〉, 〈ξ0, …, ξk〉, θ].
In this chapter, the availability concepts that are needed for the calculus are established. Our course of action can be sketched as follows: First, preliminary concepts concerning segments and segment sequences are to be established, where a segment in a sentence sequence will be a non-empty, uninterrupted subset of (2.1). Second, closed seg- ments will be characterised as certain CdI-, NI- and RA-like segments, i.e. certain seg- ments of the kinds that are connected to inferences by conditional introduction (CdI), ne- gation introduction (NI) and particular-quantifier elimination (PE) (2.2). The availability concepts themselves will then be established with recourse to closed segments. This will be done in such a way that exactly those propositions are available in a sentence sequence at a position that do not lie within a proper initial segment of a closed segment in this sen- tence sequence at this position (2.3). With the theorems that are established in this chap- ter, we can later show that CdI, NI and PE and only CdI, NI and PE can discharge as- sumptions.
2.1 Segments and Segment Sequences
First, segments in a non-empty sequence will be characterised as non-empty and unin- terrupted subsets of . Second, some theorems on segments will be proved. Then, some concepts and theorems concerning segment sequences for sentence sequences will be established, where a segment sequence for a sentence sequence is a finite sequence that enumerates disjunct segments in . Then, AS-comprising segment sequences for seg- ments in sentence sequences will be defined with recourse to segment sequences. An AS- comprising segment sequence for a segment in will be a segment sequence for for which it holds that all values of the sequence are disjunct subsegments of and that all assumption-sentences in lie in one of the values of the sequence. These AS-comprising segment sequences will later play a crucial role in the inductive generation of closed seg- ments. The end of the chapter contains the proofs of theorems about AS-comprising seg- ment sequences that are needed for the establishment of closed segments and of theorems on these. We start with the segment definition:
Definition 2-1. Segment in a sequence (metavariables: , , , ', ', ', *, *, *, …)
is a segment in iff
∈ SEQ, ≠∅, ⊆ and = {(i, i) | min(Dom( )) ≤i≤ max(Dom( ))}.