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Proof step 5 from proof contains-rec-01 (conclusion)

⊦ 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) ∧ tys .first ≤ fs .type)) ∧ (is_arraytype(ty1) → tys .first ≤ ty1 .type) ∧ contains(str, tys)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ 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) ∧ tys .first ≤ fs .type)) ∧ (is_arraytype(ty1) → tys .first ≤ ty1 .type) ∧ contains(str, tys))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]