Proof of the Detailed Soundness Lemma

We are now ready to present the proof of the Detailed Soundness Lemma. We prove this lemma by induction over the complexity of θ. So we fix aPTLtext θ0 and assume that the lemma holds for all subtexts of θ0. Additionally, we assume the following properties:

[i] Tis a list ofPTL-PLterms such that PL⁻¹(T)⊕qt(θ0) is pairwise inde-pendent.

[ii] All MHF terms ofθ0 are composed of terms inPL⁻¹(T).

[iii] Γ is a premise list such that all MHF terms in Γ are composed of PTL_sk symbols and terms inT.

[iv] check text(θ₀,Γ,T, µ) = (Γ⁰,T⁰, ν).

[v] M is a CMTN model, S a Γ-skolem-assignment and g an M-assignment such thatM +S, g|= Γ.

[vi] dom(g) =PL⁻¹(T).

[vii] For allT ∈T, ^M+S_g (T)6=u.

Now we distinguish 13 different cases depending on the form ofθ₀. Case 1: θ0 is aPTL term t

In this case, there is a premise list Γ^∗, aPLtermT and a proof status valueµ0

such that the following hold:

(i) read term(t,Γ,T, µ) = (Γ^∗, T, µ0).

(ii) µ1=update(µ0,0, P(Γ^∗`^?B(T))).

(iii) ν =update(µ₁,1, P(Γ^∗⊕ hB(T)i `^?T =>)).

(iv) Γ⁰= Γ^∗⊕ hB(T)^P, T =>i.

Now we prove each of the six assertions of the Detailed Soundness Lemma fort:

1. By the definition of read term, T⁰−T contains noD-marked terms, i.e.

PL⁻¹(T⁰−T) =∅=tbc(t), as required.

2. This directly follows from assertions 1 and 2 of the read termSoundness Lemma.

3. Left-to-right implication:

Suppose that def(JtK

g

M). Now by the definition of def(JtK

g

M),B^M(^M_g (t)), which by the Sort Disjointness Axiom ofCMTNimplies that ^M_g(t)6=u^M. By assertion 3 of theread termSoundness Lemma and using the fact that T⁰−T=∅, pres(Γ⁰,Γ,T⁰−T, M, S, g,Φ) holds for all presuppositionally marked Φ in Γ^∗−Γ. Now we still have to show that pres(Γ⁰,Γ,T⁰ − T, M, S, g, B(T)). So let S⁰ S be a Γ⁰-skolem-assignment such that

6.3. SOUNDNESS OF THE PROOF CHECKING ALGORITHM 131 M +S⁰, g|= Γ^∗. Now by assertion 4 of the read termSoundness Lemma,

M

g(t) =^M+S_g ⁰(T), i.e.B^M(^M+S_g ⁰(T)), i.e.M+S⁰, g|=B(T), as required.

Right-to-left implication:

Suppose that for all presuppositionally marked premises Φ in Γ⁰ −Γ, pres(Γ⁰,Γ,T⁰ −T, M, S, g,Φ). Note that by the definition of read term, Γ^∗−Γ contains only presuppositionally marked premises. This allows us to conclude that there is a Γ^∗-skolem-assignment S₀ S such that M+S0, g|= Γ^∗−Γ. Now sincepres(Γ⁰,Γ,T⁰−T, M, S, g, B(T)),M+S0, g|= B(T), i.e. B^M(^M+S_g ⁰(T)). Note that by assertion 3 of the read term Soundness Lemma, ^M_g(t) 6= u^M. Now by assertion 4 of the read term Soundness Lemma, ^M_g(t) = ^M+S_g ⁰(T), i.e. B^M(^M_g(t)), i.e. def(JtK

g M), as required.

4. Suppose thatdef(JtK

g

M), and letk be anM-assignment.

(a)⇒(b):

Suppose k ∈ JtK

g

M. Then k = g, i.e. k[T⁰−T]g. Now it is enough to show that g verifies Γ⁰−Γ over M+S. For this suppose thatS⁰ S is a Γ^∗-skolem-assignment such that M+S⁰, g|= Γ^∗−Γ. We have to show that M+S⁰, g|=T =>.

SinceJtK

g

M 6=∅, ^M_g(t) =>^M. But by assertion 4 of theread term Sound-ness Lemma, ^M_g (t) =^M+S_g ⁰(T), so M+S⁰, g|=T =>, as required.

(b)⇒(c):

This implication from (b) to (c) actually does not depend on which of the 13 cases of this proof we are in. So we only present the proof for this implication once here, and leave it out in all later cases.

Suppose thatk[T⁰−T]g and thatkverifies Γ⁰−Γ overM+S. We prove inductively that for every initial segment Γ0 of Γ⁰, there is a Γ0 -skolem-assignmentS⁰ S such thatM +S⁰, k|= Γ₀.

For Γ₀= Γ, we can deduceM+S, k|= Γ from the fact thatM+S, g|= Γ and the fact that no free term in Γ is inT⁰−T(which intuitively follows from the fact that the proof checking algorithm only keeps track of terms that have occurred in the already processed parts of a PTL text, and which can be proved formally from the fact that tbc(θ0) =T⁰−T, from our basic assumptions [i] and [iii] and from the definition oftbc).

For the inductive step, suppose that Φ is a premise in Γ and thatS⁰ Sis a ΓΦ-skolem-assignment such thatM+S⁰, k|= ΓΦ. It now suffices to show that there is a ΓΦ+-skolem-assignmentS⁰⁰S⁰such thatM+S⁰⁰, k|= Φ.

If Φ is not presuppositionally marked, thenM+S⁰, k|= Φ follows directly from the fact thatkverifies Γ⁰−Γ overM+S. So suppose Φ is presuppo-sitionally marked. Sincedef(Jθ0K

g

M), the already proved assertion 3 of the Detailed Soundness Lemma forθ0implies thatpres(Γ⁰,Γ,T⁰−T, M, S, g,Φ).

This now implies the required result that there is a Γ_Φ+-skolem-assignment S⁰⁰S⁰ such thatM +S⁰⁰, k|= Φ.

(c)⇒(a):

Suppose thatk[T⁰−T]g and that there is a Γ⁰-skolem-assignmentS⁰ S such that M +S⁰, k |= Γ⁰. Since T⁰−T=∅, k=g. ThenM +S⁰, k |=

T =>, i.e. ^M+S_k ⁰(T) =>^M. Sincedef(JtK

g

M), ^M_k(t)6=u^M, so by assertion 4 of the read termSoundness Lemma, ^M_k(t) = ^M+S_k ⁰(T) =>^M. Now by the definition of JtK

g

M,k=g∈JtK

g

M, as required.

5. It is enough to show that if ν 6=u, then µ+v(t, M, g) 6=u, and that if ν =>, thenµ+v(t, M, g) =>.

So suppose ν 6= u. By (iii), µ1 6= u, i.e. by (ii), µ0 6= u and CMTN∪ Γ^∗ |=B(T), i.e. pres(Γ⁰,Γ,T, M, S, g, B(T)). Then by assertion 5 of the read termSoundness Lemma,µ6=uand^M_g(t)6=u. So by the above estab-lished assertion 3 of this lemma, v(t, M, g)6=u. Henceµ+v(t, M, g)6=u.

Now suppose ν =>. Then by (iii), µ₁=>and CMTN∪Γ^∗⊕ hB(T)i |= T =>. So by (ii),µ₀=>. Now by assertion 5 of theread termSoundness Lemma, µ=> and ^M_g(t)6=u, i.e. def(JtK

g

M). In order to conclude that µ+v(t, M, g) = >, it is now enough to show that v(t, M, g) = >, i.e.

that JtK

g

M 6=∅. By the above established assertion 4 of this lemma, it is therefore enough to show thatgverifies Γ⁰−Γ overM+S. So letS⁰ S be a Γ^∗⊕ hB(T)i-skolem-assignment such thatM+S⁰, g|= Γ^∗⊕ hB(T)i.

SinceCMTN∪Γ^∗⊕ hB(T)i |=T =>, M+S⁰, g|=T =>, as required.

6. Trivial (sinceT⁰=T).

Case 2: θ₀ is of the formR(t₁, . . . , t_n)

In this case there are premise lists Γ₁, . . . ,Γ_n+1, term lists T1, . . . ,Tn+1 and proof status valuesµ₁, . . . , µ_n+1 such that the following properties hold:

(i) Γ₁= Γ.

(ii) T1=T. (iii) µ1=µ.

(iv) for all 1≤i≤n, read term(ti,Γi,Ti, µi) = (Γi+1, Ti, µi+1).

(v) Γ⁰= Γ_n+1⊕ hR(T1, . . . , T_n)i.

(vi) T⁰ =Tn+1.

(vii) ν =update(µn+1,1, P(Γn+1`^?R(T1, . . . , Tn))).

Now we prove each of the six assertions of the Detailed Soundness Lemma forR(t1, . . . , tn):

1. By the definition ofread termand (iv),PL⁻¹(T⁰−T) =∅=tbc(R(t₁, . . . , t_n)).

2. This follows directly from assertion 2 of theread termSoundness Lemma applied to all applications of read termin (iv).

3. This follows from the fact thatdef(JR(t1, . . . , tn)K

g

M) iff for all 1≤i≤n,

M

g (ti) 6= u^M, and from assertion 3 of the read term Soundness Lemma applied to all applications of read termin (iv).

6.3. SOUNDNESS OF THE PROOF CHECKING ALGORITHM 133 4. Assume def(JR(t1, . . . , tn)K

g

M). Then for all 1≤i≤n, ^M_g (ti)6=u^M. (a)⇒(b):

Assume k ∈JR(t₁, . . . , t_n)K

g

M. Thenk=g, i.e.k[T−T]g. It remains to be shown that k verifies Γ⁰−Γ overM +S. For this, let S⁰ S be a Γ_n+1-skolem-assignment such thatM +S⁰, g|= Γ_n+1. Now we only need to show thatM+S⁰, g|=R(T₁, . . . , T_n).

Since g∈JR(t1, . . . , tn)K

g

M, (^M_g(t1), . . . ,^M_g(tn))∈R^M. By assertion 4 of theread termSoundness Lemma, ^M_g(t_i) = ^M+S_g ⁰(T_i) for 1≤i≤n. Now (^M+S_g ⁰(T₁), . . . ,^M+S_g ⁰(T_n))∈R^M, i.e.M +S⁰, g|=R(T₁, . . . , T_n).

(b)⇒(c):

As in case 1.

(c)⇒(a):

Let S⁰ S be a Γ⁰-skolem-assignment such that M +S⁰, g |= Γn+1 ⊕ hR(T1, . . . , Tn)i. By assertion 4 of the read term Soundness Lemma,

M

g(ti) = ^M+S_g ⁰(Ti) for 1 ≤ i ≤ n, so (^M_g (t1), . . . ,^M_g (tn)) ∈ R^M, i.e.

g∈JR(t₁, . . . , t_n)K

g

M, as required.

5. Suppose ν 6= u. Then by (vii), µn+1 6= u. So by assertion 5 of the read term Soundness Lemma,µn =· · · =µ1 6=u and for all 1≤i≤n,

M

g(ti)6=u^M. This implies thatµ=µ16=uand thatdef(JR(t1, . . . , tn)K

g M), i.e.µ+v(R(t1, . . . , tn), M, g)6=u.

Now suppose ν = >. Then by (vii), µ_n+1 = > and CMTN∪Γ_n+1 |= R(T₁, . . . , T_n). So by assertion 5 of the read term Soundness Lemma, µ_n = · · · = µ₁ = > and for all 1 ≤ i ≤ n, ^M_g(t_i) = >. Now it follows that µ = µ1 = > and def(JR(t1, . . . , tn)K

g

M). In order to con-clude thatµ+v(R(t1, . . . , tn), M, g) =>, it is now enough to show that v(R(t1, . . . , tn), M, g) =>, i.e. that JR(t1, . . . , tn)K

g

M 6=∅. By the above established assertion 4 of this lemma, it is therefore enough to show that g verifies Γ⁰−Γ over M+S. So letS⁰S be a Γ_n+1-skolem-assignment such that M +S⁰, g |= Γ_n+1. Since CMTN∪Γ_n+1 |= R(T₁, . . . , T_n), M +S⁰, g|=R(T₁, . . . , T_n), as required.

6. Trivial.

Case 3: θ0 is of the form ¬ϕ

In this case there are Γ^∗,T0, Φ andµ⁰ such that the following hold:

(i) read text(ϕ,hi,Γ,T, µ) = (Γ^∗,T0,Φ, µ⁰).

(ii) ν =update(µ⁰,1, P(Γ^∗`^?¬∃_T0 Φ)).

(iii) Γ⁰ = Γ^∗⊕ h¬∃_T0 Φi.

(iv) T⁰=T.

Now by theread textSoundness Lemma, the following properties hold:

I. All MHF terms of Φ are composed ofPTLsksymbols and terms inT∪T0.

II. All MHF terms in Γ^∗−Γ are composed ofPTLsksymbols and terms inT. III. Eitherµ⁰=uorµ⁰=µanddef(JϕK

g M).

IV. def(JϕK

g

M) iff for every Ψ in Γ^∗−Γ and every Γ^∗_Ψ-skolem-assignmentS⁰ S such that M +S⁰, g|= Γ^∗_Ψ, there is a Γ^∗-skolem-assignmentS0 S⁰ such thatM +S0, g|= Γ^∗.

V. Ifdef(JϕK

g

M), then the following three properties are equivalent:

(a) k∈JϕK

g M.

(b) k[T0]g and for every Γ^∗-skolem-assignment S⁰ extendingS such that M+S⁰, g|= Γ^∗, M+S⁰, k|= Φ.

(c) k[T0]g and there is a Γ^∗-skolem-assignmentS⁰ extendingS such that M+S⁰, k|= Γ^∗⊕ hΦi.

Now we prove the six assertions of the Detailed Soundness Lemma for¬ϕ:

1. tbc(¬ϕ) =∅=PL⁻¹(T−T) as required.

2. It easily follows from I and II that all MHF terms in Γ⁰−Γ are composed ofPTL_sksymbols and terms inT, as required.

3. def(J¬ϕK

g

M) iffdef(JϕK

g M)

iff for every Ψ in Γ^∗−Γ and every Γ^∗_Ψ-skolem-assignment S⁰ S such that M +S⁰, g|= Γ^∗_Ψ, there is a Γ^∗-skolem-assignmentS0 S⁰ such that M +S0, g|= Γ^∗ (by assertion IV)

iff for every presuppositionally marked premise Ψ in Γ⁰−Γ, every Γ⁰_Φ -skolem-assignmentS⁰S and everyk[T−T]gsuch thatM+S⁰, k|= Γ⁰_Ψ, there is a Γ^∗-skolem-assignmentS₀S⁰ such thatM+S₀, g|= Γ^∗ (since

¬∃_TΦ is not presuppositionally marked in Γ⁰), as required.

4. Assume def(J¬ϕK

g

M). Thendef(JϕK

g M).

(a)⇔(c):

k∈J¬ϕK

g

M iffk=g and there is nok⁰∈JϕK

g M

iffk=gand there is nok⁰[T0]g such that for every Γ^∗-skolem-assignment S⁰ extendingSsuch thatM+S⁰, g|= Γ^∗,M+S⁰, k⁰|= Φ (by the equiva-lence of (a) and (b) in V)

iff k = g and there is a Γ^∗-skolem-assignment S⁰ extending S such that M+S⁰, g|= Γ^∗ and such that there is nok⁰[T0]gsuch thatM+S⁰, k⁰|= Φ iff k =g and there is a Γ⁰-skolem-assignment S⁰ extendingS such that M +S⁰, g |= Γ⁰ (note that a Γ⁰-skolem-assignment is just a Γ^∗ -skolem-assignment, since ¬∃_T0 Φ does not introduce skolem functions), as re-quired.

(a)⇔(b):

k∈J¬ϕK

g

M iffk=g and there is nok⁰∈JϕK

g M

iffk=gand there is nok⁰[T0]gand no Γ^∗-skolem-assignmentS⁰extending S such thatM +S⁰, k⁰ |= Γ^∗⊕ hΦi(by the equivalence of (a) and (c) in V)

6.3. SOUNDNESS OF THE PROOF CHECKING ALGORITHM 135 iffk=gand for every Γ^∗_Φ-skolem-assignmentS⁰Ssuch thatM+S⁰, g|= Γ^∗−Γ, there is no k⁰[T0]g such thatM +S⁰, k⁰ |= Φ (by II and the fact that M+S⁰, g|= Γ forS⁰ S)

iffk=gand for every Γ^∗_Φ-skolem-assignmentS⁰Ssuch thatM+S⁰, g|= Γ^∗−Γ, we haveM+S⁰, g|=¬∃_T0 Φ

iffk=g andkverifies Γ^∗−Γ overM+S, as required.

5. Supposeν 6=u. Thenµ⁰ 6=u, i.e.µ6=uanddef(JϕK

g

M) by III. But then def(J¬ϕK

g

M), i.e.v(¬ϕ, M, g)6=u, i.e.µ+v(¬ϕ, M, g)6=u.

Now suppose ν =>. Thenµ⁰ => andP(Γ^∗ `^? ¬∃_T0 Φ) = >. This on the one hand implies µ =>by III, and on the other hand implies that CMTN∪Γ^∗ |=¬∃_T0 Φ. But then by the just proved assertion 4 of the Detailed Soundness Lemma for ¬ϕ, g ∈ JϕK

g

M, i.e. v(¬ϕ, M, g) = >, i.e.

µ+v(¬ϕ, M, g) =>.

6. Trivial.

Case 4: θ0 is of the form θ∧ψ

This case can be verified in a way very similar to case 3.

Case 5: θ0 is of the form ϕ∨ψ

This case can be verified in a way very similar to case 3.

Case 6: θ₀ is of the form ϕ→θ

In this case, there are Γ0, Γ1, Γ2, Γfunc,T0,T1,T2,T^∗,F, Φ,α,µ0andµ1such that the following hold:

(i) read text(ϕ,hi,Γ,T, µ) = (Γ₀,T1,Φ, µ₀).

(ii) check text(θ,Γ₀⊕ hΦi,T⊕T1, µ₀) = (Γ₁,T2, µ₁).

(iii) α=









 1

if the symbol L does not occur in Γ1−Γ0 and for every term T occurring in Γ1−Γ0 that is either in Tor a skolem function symbol, check limitedness(Γ₁,Γ₁−Γ₀,T, T)

0 otherwise.

(iv) make functions(T1,T2−(T⊕T1),Γ₀,Γ₁,Φ, α, µ₁) = (F,T^∗,Γ_func,Γ⁻_pres, ν).

(v) pull out pres(hi,T2−T,Γ₀,Γ₁) = (Γ^∗,h i ⊕Γ₂, ).

(vi) Θ =∃_T^∗V Γ2.

(vii) Γpres= Γ⁻_pres⊕ h∀_T2−T(Φ∧Θ→L(T))^P |T ∈T2−(T⊕T1)−T^∗i.

(viii) Γ⁰ = Γ^∗⊕Γpres⊕ h∀_T1 (Φ→Θ)i ⊕Γfunc. (ix) T⁰=T⊕F.

Note that the assertions (ii), (v) and (vi) closely resemble the definition of read text. Using arguments analogous to those in the proof of the read text Soundness Lemma, we can derive from them the following assertions:

I. tbc(θ) =T2−(T⊕T1).

II. All MHF terms ofVΓ2 are composed ofPTLsk symbols and terms inT2. III. All MHF terms in Γ^∗−Γ0 are composed of PTLsk symbols and terms in

T.

IV. Suppose thatdef(JϕK

g

M). Then the following two properties are equivalent:

(a) For allh∈JϕK

g

M,def(JθK

h M).

(b) For every Ψ in Γ^∗−Γ₀ and every Γ^∗_Ψ-skolem-assignmentS⁰ S such that M +S⁰, h|= Γ^∗_Ψ, there is a Γ^∗-skolem-assignmentS₀ S⁰ such thatM+S0, h|= Γ^∗.

V. Suppose that h∈JϕK

g

M, that def(JθK

h

M) and that S⁰ S is a Γ₀ -skolem-assignment such that M +S⁰, h |= Γ₀ ⊕ hΦi. Then the following three properties are equivalent:

(a) k∈JθK

h M.

(b) k[T2−(T⊕T1)]h and for every Γ^∗-skolem-assignment S⁰⁰ extending S⁰ such thatM+S⁰⁰, h|= Γ^∗,M +S⁰⁰, k|= Γ2.

(c) k[T2−(T⊕T1)]hand there is a Γ^∗-skolem-assignment S⁰⁰ extending S⁰ such thatM+S⁰⁰, k|= Γ^∗⊕Γ2.

Now we prove the six assertions of the Detailed Soundness Lemma forϕ→θ:

1. By the definition ofmake function, PL⁻¹(T⁰−T) =PL⁻¹(F)

=PL⁻¹({T⁰ |for some T in T2 − (T ⊕T1), there is a length(T1)-place argument fillerσ such that T =T^0σ(T1)})

=tbc(ϕ→θ)

by the definition of aq and tbc and the facts that tbc(ϕ) = PL⁻¹(T1) (assertion 1 of the read text Soundness Lemma) and that tbc(θ) = PL⁻¹(T2−(T⊕T1)) (assertion 1 of the Detailed Soundness Lemma).

2. Let T be an MHF term in Γ⁰−Γ. We have to show thatT is composed ofPTLsksymbols and terms inT⊕F. For this we distinguish four cases:

Case 1: T is an MHF term in Γ0−Γ.

In this case,T is composed ofPTLsksymbols and terms inTby assertion 3 of the read textSoundness Lemma.

Case 2: T is an MHF term in Γ^∗−Γ0.

In this case,T is composed ofPTLsksymbols and terms inTby assertion III above.

Case 3: T is an MHF term in Γpres. This case is similar to case 4 below.

6.3. SOUNDNESS OF THE PROOF CHECKING ALGORITHM 137

Im Dokument Proof-checking mathematical texts in controlled natural language (Seite 144-151)