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Proof step 4 from proof contains-rec-01 (previous step)

⊦ 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 elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
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))

Rule argument

[speclemmaarg: tys ≠ [] ,ty = tys .first ∧ tys0 = tys .rest ↔ tys = ty + tys0 ,[tys, ty, tys0] → [tys, ty0, tys0] ,T ,list-data ,javatypes ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : tys ≠ [] ⊦ ty = tys .first ∧ tys0 = tys .rest ↔ tys = ty + tys0])], [EXTRATERMS : (tys, ty0, tys0)]]