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Proof step 19 from proof contains-append (previous step)

∃ 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), tys1 + ty ' = ty2 ' + tys, is_reftype(ty0), type∃(str, ty0), ∀ str, ty, tys0. contains(str, tys1 + ty ' + tys0) ↔ contains(str, tys1 + ty ') ∧ contains(str, ty ' + tys0) ⊦ (∃ 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, ty2 ' + tys)) ∧ contains(str, ty ' + tys0), ¬ ty2 ∈ tys1 ∧ ty2 ≠ ty, ty ≠ ty2 ∧ ¬ ty ∈ tys

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
tys1 + ty ' = ty2 ' + tys, ty1 ≤ ty0, type∃(str, ty1), contains(str, tys1 + ty ' + tys0), is_reftype(ty0), type∃(str, ty0), is_arraytype(ty1) → ty2 ≤ ty1 .type, ∀ str, ty, tys0. contains(str, tys1 + ty ' + tys0) ↔ contains(str, tys1 + ty ') ∧ contains(str, ty ' + tys0), is_classtype(ty1) → ¬ interface∃(str, ty1 .class) ∧ fs ' ∈ instfields(ty1 .class) ∧ ty2 ≤ fs .type ⊦ (∃ 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, ty2 ' + tys)) ∧ contains(str, ty ' + tys0), ¬ ty2 ∈ tys1 ∧ ty2 ≠ ty, ty ≠ ty2 ∧ ¬ ty ∈ tys

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]