The rules of G4ip are more complex than those of G3ip. They are also more numerous. Whereas all Cut admissibility for G3ip needed was Weakening and invertibility, here we require further results before Contraction can be shown admissible. Before these extra results, however, we note the following invertibility lemmata which areused4:
Lemma 1 (Invertibility of G4ip [Dyckhoff and Negri, 2000]) The following rules are height-preserving admissible in G4ip:
Γ, A∨B ⇒E Γ, A⇒E L∨1,inv Γ, A∨B ⇒E Γ, B⇒E L∨2,inv Γ, A∧B ⇒E Γ, A, B⇒E L∧inv Γ⇒A⊃B Γ, A⇒B R⊃inv Γ,At i⊃B⇒E Γ, B⇒E L0⊃inv Γ,(C∧D)⊃B ⇒E Γ, C⊃(D⊃B)⇒E L∧ ⊃inv Γ,(C∨D)⊃B⇒E Γ, C⊃B, D⊃B⇒E L∨ ⊃inv Γ,(C⊃D)⊃B⇒E Γ, B⇒E L⊃⊃inv
The proofs are standard inductions.
As for the non-standard lemmata, firstly we show that generalised axioms are admissible5:
Γ, A⇒A
for arbitraryA. This is done by an strong induction on the weight ofA:
lemmagenAx:
assumesw =weight A shows∃ n.Γ⊕A⇒A↓n usingassms
proof (induct w arbitrary:A rule:nat-less-induct)
WhereAis a propositional atom,⊥, a conjunction or a disjunction, the result is quite simple. Where the main connective is an implication, we perform case analysis on the antecedent:
case(Imp E F)
then show∃ n.Γ⊕A⇒A↓n using hA=E⊃Fi
proof (cases E)
4R∧is invertible, but this is not required for the formal proofs which follow. 5Other authors call them expanded axioms e.g. [Ciabattoni and Terui,2006a].
This gives five cases: • A= Ati ⊃F • A=⊥ ⊃F • A= (G∧H)⊃F • A= (G∨H)⊃F • A= (G⊃H)⊃F
Only four of these cases are treated in [Dyckhoff and Negri,2000], although the missing case (A=⊥ ⊃F) is trivial. We show the most complicated case, whereA= (G∧H)⊃F:
Γ, G⊃(H⊃F)⇒G⊃(H⊃F) IH Γ, G⊃(H⊃F), G⇒H⊃F inv Γ, G⊃(H⊃F), G, H ⇒F inv Γ,(G∧H)⊃F, G, H ⇒F L∧ ⊃ Γ,(G∧H)⊃F, G∧H ⇒F L∧ Γ,(G∧H)⊃F ⇒(G∧H)⊃F R⊃
As noted in [Dyckhoff and Negri, 2000], the induction hypothesis applies because w(G⊃
(H⊃F))< w((G∧H)⊃F). This is formalised as follows: case(Conj G H)
then haveweight (G⊃(H⊃F))<weight Ausing hA=E⊃Fi byauto
then have∃ n.Γ⊕G⊃(H⊃F)⇒G⊃(H⊃F)↓n usingIH byauto
then have∃ n.Γ⊕G⊃(H⊃F)⊕G ⇒H⊃F ↓n usinginversionImpR byauto then have∃ n.Γ⊕G⊃(H⊃F)⊕G ⊕H ⇒F ↓n usinginversionImpR byauto then obtainn wherea: Γ ⊕G⊃(H⊃F) ⊕G ⊕H ⇒F ↓nbyblast
froma haveΓ⊕(G∧∗H)⊃F ⊕G⊕H ⇒F ↓n+1 usingprovable.ImpLC by(auto simp add:union-ac)
then haveΓ⊕(G∧∗H)⊃F ⊕G∧∗H ⇒F ↓n+2 usingprovable.ConjLby(auto simp add:union-ac) then haveΓ⊕(G∧∗H)⊃F ⇒(G∧∗H)⊃F ↓n+2+1
usingprovable.ImpRbyauto
then show∃ n.Γ⊕A⇒A↓n usingprems byauto
That there is a case not covered is surprising. However, it shows the usefulness of a for- malisation, since it was formalisation which uncovered the missing case. Had this result not been formalised the gap, however small, may have remained:
caseff
then haveΓ⊕ff ⊃F ⇒ff ⊃F ↓1 usingprovable.ImpRbyauto then show∃ n.Γ⊕A⇒A↓n byauto
Using this lemma, it is easy to show that modus ponens is derivable:
Γ, A⊃B⇒A⊃B
Γ, A, A⊃B⇒B R⊃inv
The next auxiliary lemma required is the following:
Lemma 2 The rule:
Γ⇒D Γ, B⇒E Γ, D⊃B⇒E
is admissible inG4ip.
Proof. Induction on the heightnof the derivation of the left premiss. Ifn= 0, there are two subcases. In the first,⊥ ∈Γ, and so the conclusion is an instance ofL⊥. In the second case, there is some propositional atom Atisuch thatD = Atiand Ati∈Γ. We can then applyL0⊃ to the second premiss.
Whenn >0 there are several cases:
1. The last inference is an invertible left inference: L∧, L∨, L∧ ⊃, L∨ ⊃, L0⊃. Then, we apply the appropriate inversion lemma to the right premiss, the induction hypothesis, and then the appropriate rule.
2. The last inference is a right rule: R∧, R∨, R⊃. In each case, a combination of the induction hypothesis on the premiss(es), Weakening and the appropriate rule from L∧ ⊃, L∨ ⊃, L⊃⊃ is used.
3. The last inference was an instance of L⊃⊃. Then, Γ = Γ0⊕(F⊃G)⊃H and the premisses are Γ0⊕G⊃H⊕F ⇒Gand Γ0⊕H⇒E. The following derivation suffices:
Γ0, G⊃H, F ⇒G Γ0, G⊃H, F, E⊃B ⇒G w Γ0, H⇒E Γ0,(F⊃G)⊃H, B⇒S Γ0, H, B⇒S inv Γ0, H, E⊃B⇒S ih Γ0,(F⊃G)⊃H, E⊃B⇒S L⊃⊃ a
The formalisation of this case is as follows:
case(ImpLLΓ0F G H n E m)
then obtaina whereΓ0⊕(F⊃G)⊃H ⊕B ⇒S ↓a byblast
then havea:∃a.Γ0⊕H ⊕E⊃B ⇒S ↓a using prems(6)[whereB=B]byauto
moreover— Left hand derivation
fromhΓ0⊕F ⊕G⊃H ⇒G ↓nihave Γ0⊕G⊃H ⊕F ⊕E⊃B ⇒G ↓n
usingdpWeak by(auto)
ultimately show?case usingprovable.ImpLLby(auto simp add:union-ac)
The final auxiliary lemma we need is one which is somewhat counter-intuitive:
Lemma 3 (Double D⊃B) The rule:
Γ,(C⊃D)⊃B⇒E Γ, C, D⊃B, D⊃B⇒E
is admissible inG4ip.
Given that we are trying to showContraction is admissible, showing this seems unhelpful. However, D⊃B is a lighter formula than (C⊃D)⊃B, so we can use the manipulation to create two lighter formulae. In the proof, when (C⊃D)⊃B is non-principal, the result is straightforward. The proof is by induction on the height of the derivation of the premiss, and the interesting case is where the last inference has (C⊃D)⊃B principal. Then, the premisses are Γ⊕C⊕D⊃B⇒D and Γ⊕B⇒E and the following is a valid derivation:
Γ, C, D⊃B⇒D
Γ, B⇒E
Γ, C, D⊃B, B⇒E w Γ, C, D⊃B, D⊃B⇒E Lemma2
This is formalised as follows:
lemmatwoDB:
assumesΓ⊕(C⊃D)⊃B ⇒E ↓n
shows∃ m.Γ⊕C ⊕D⊃B ⊕D⊃B ⇒E ↓m usingassms
proof (induct Γ≡Γ⊕(C⊃D)⊃B E n arbitrary: Γ)
case(ImpLLΓF G H n E m Γ0)
have(F⊃G)⊃H = (C⊃D)⊃B ∨(F⊃G)⊃H 6= (C⊃D)⊃B byblast
moreover
{assume(F⊃G)⊃H = (C⊃D)⊃B then haveeqs:F=C
G=D H=B
Γ=Γ0usingprems byauto
moreover
haveΓ⊕F ⊕G⊃H ⇒G↓n byfact ultimately
have∃ n.Γ⊕F ⊕G⊃H ⊕G⊃H ⇒E ↓n using ImpLClassical by(auto simp add:union-ac)
then have∃ n.Γ0⊕C ⊕D⊃B ⊕D⊃B ⇒E ↓nusing eqsbyblast
}
moreover
{assume(F⊃G)⊃H 6= (C⊃D)⊃B— Non-principal case. Proof omitted have∃ n.Γ0⊕C ⊕D⊃B ⊕D⊃B ⇒E ↓nbyblast
}
ultimately
show?case byblast