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Proof step 6 from proof contains-trans (previous step)

contains(str, ty0 ' + tys + ty1 '), contains(str, ty ' + tys0 + ty0 ') ⊦ dups(tys), ty0 ∈ tys, dups(tys0), ty ∈ tys0, contains(str, ty ' + (tys0 + (ty0 + tys)) + ty1 ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
contains(str, ty0 ' + tys + ty1 '), contains(str, ty ' + tys0 + ty0 ') ⊦ dups(tys), ty0 ∈ tys, dups(tys0), ty ∈ tys0, contains(str, ty ' + tys0 + (ty0 ' + tys) + ty1 ')

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ty + tys = ty ' + tys])]]