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Proof step 71 from proof contains-levelsubset (previous step)

tys + ty0 ' ≠ [], 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 + ty0 ') .first ≤ fs .type)) ∧ (is_arraytype(ty1) → (tys + ty0 ') .first ≤ ty1 .type) ∧ contains(str, tys + ty0 ')), level⊆(str, str0), ∀ ty0, ty, str0, str, tys0. # tys0 < # tys → (level⊆(str, str0) ∧ contains(str, ty ' + tys0 + ty0 ') → contains(str0, ty, ty0) ∨ dups(tys0) ∨ ty ∈ tys0) ⊦ ty ∈ tys, dups(tys), contains(str0, ty, ty0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
type∃(str, ty), is_reftype(ty), ty1 ≤ ty, type∃(str, ty1), contains(str, tys + ty0 '), level⊆(str, str0), is_arraytype(ty1) → (tys + ty0 ') .first ≤ ty1 .type, is_classtype(ty1) → ¬ interface∃(str, ty1 .class) ∧ fs ' ∈ instfields(ty1 .class) ∧ (tys + ty0 ') .first ≤ fs .type, ∀ ty0, ty, str0, str, tys0. # tys0 < # tys → (level⊆(str, str0) ∧ contains(str, ty ' + tys0 + ty0 ') → contains(str0, ty, ty0) ∨ dups(tys0) ∨ ty ∈ tys0) ⊦ ty ∈ tys, dups(tys), contains(str0, ty, ty0)

Rule argument

[emptyarg]

Additional information

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