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Proof step 3 from proof contains-rec-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ contains(str, ty ' + ty0 ' + tys0) ↔ type∃(str, ty) ∧ is_reftype(ty) ∧ (∃ ty1. ty1 ≤ ty ∧ type∃(str, ty1) ∧ (is_classtype(ty1) → ¬ interface∃(str, ty1 .class) ∧ (∃ fs. fs ' ∈ instfields(ty1 .class) ∧ ty0 ≤ fs .type)) ∧ (is_arraytype(ty1) → ty0 ≤ ty1 .type) ∧ contains(str, ty0 ' + tys0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (tys + tys0) + tys1 = tys + tys0 + tys1], [CSIMPSEQ : ⊦ ty ' + tys ≠ []], [CSIMPSEQ : ⊦ ty + tys = ty ' + tys])]]