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Proof step 5 from proof delete-dups (previous step)

dups((a0 ' + x0) -l a), ∀ a1. ¬ dups(x0 -l a1) ⊦ a = a0, a0 ∈ x0, dups(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
dups(x0 -l a), ∀ a1. ¬ dups(x0 -l a1) ⊦ a = a0, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);%in-del-new: ⊦ a ∈ x -l b ↔ a ∈ x ∧ a ≠ b;%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list