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Proof step 10 from proof contains-trans (previous step)

∃ tys. contains(str, ty ' + (tys + ty0)) ∧ ¬ dups(ty ' + tys), contains(str, ty0 ' + tys + ty1 ') ⊦ contains(str, ty, ty1), dups(tys), ty0 ∈ tys

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
contains(str, ty0 ' + tys + ty1 '), contains(str, ty ' + tys0 + ty0 ') ⊦ contains(str, ty, ty1), dups(tys), ty0 ∈ tys, dups(tys0), ty ∈ tys0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (tys + tys0) + tys1 = tys + tys0 + tys1], [CSIMPSEQ : ⊦ dups(ty ' + tys) ↔ dups(tys) ∨ ty ∈ tys], [CSIMPSEQ : ⊦ tys + ty = tys + ty '])]]