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Proof step 35 from proof contains-append (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
tys1 + ty ' = ty2 ' + tys, ∀ 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), ty ≠ ty2 ∧ ¬ ty ∈ tys, ¬ ty2 ∈ tys1 ∧ ty2 ≠ ty

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ty ∈ ty0 ' ↔ ty = ty0], [CSIMPSEQ : ⊦ ty ∈ tys + tys0 ↔ ¬ ¬ (ty ∈ tys ∨ ty ∈ tys0)], [CSIMPFORWARD : ⊦ ty ' + tys = tys0 → ty ∈ tys0], [CSIMPFORWARD : ⊦ tys + ty ' = tys0 → ty ∈ tys0], [CSIMPSEQ : ⊦ (tys + tys0) + tys1 = tys + tys0 + tys1], [CSIMPSEQ : ⊦ tys + ty ' ≠ []], [CSIMPSEQ : ⊦ ty + tys = ty ' + tys])]]