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Proof step 36 from proof contains-append (previous step)

∀ str, ty, tys0. contains(str, tys1 + ty ' + tys0) ↔ contains(str, tys1 + ty ') ∧ contains(str, ty ' + tys0) ⊦ type∃(str, ty0) ∧ is_reftype(ty0) ∧ (∃ ty1. ty1 ≤ ty0 ∧ type∃(str, ty1) ∧ (is_classtype(ty1) → ¬ interface∃(str, ty1 .class) ∧ (∃ fs. fs ' ∈ instfields(ty1 .class) ∧ (tys1 + ty ') .first ≤ fs .type)) ∧ (is_arraytype(ty1) → (tys1 + ty ') .first ≤ ty1 .type) ∧ contains(str, tys1 + ty ' + tys0)) ↔ contains(str, (ty0 ' + tys1) + ty ') ∧ contains(str, ty ' + tys0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
tys1 + ty ' = ty2 + tys ∧ tys1 + ty ' ≠ [], ∀ str, ty, tys0. contains(str, tys1 + ty ' + tys0) ↔ contains(str, tys1 + ty ') ∧ contains(str, ty ' + tys0) ⊦ type∃(str, ty0) ∧ is_reftype(ty0) ∧ (∃ ty1. ty1 ≤ ty0 ∧ type∃(str, ty1) ∧ (is_classtype(ty1) → ¬ interface∃(str, ty1 .class) ∧ (∃ fs. fs ' ∈ instfields(ty1 .class) ∧ ty2 ≤ fs .type)) ∧ (is_arraytype(ty1) → ty2 ≤ ty1 .type) ∧ contains(str, tys1 + ty ' + tys0)) ↔ contains(str, (ty0 ' + tys1) + ty ') ∧ contains(str, ty ' + tys0)

Rule argument

[speclemmaarg: tys ≠ [] ,ty = tys .first ∧ tys0 = tys .rest ↔ tys = ty + tys0 ,[tys, ty, tys0] → [tys1 + ty ', ty2, tys] ,T ,list-data ,javatypes ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ tys + tys0 ≠ [] ↔ tys = [] ∧ tys0 = []], [CSIMPSEQ : ⊦ ty ' ≠ []], [CSIMPSEQ : ⊦ (tys + tys0) + tys1 = tys + tys0 + tys1], [CSIMPELIM : tys ≠ [] ⊦ ty = tys .first ∧ tys0 = tys .rest ↔ tys = ty + tys0])], [EXTRATERMS : (tys1 + ty ', ty2, tys)]]