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Proof step 6 from proof delete-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
dups(x0 -l a), ∀ a1. ¬ dups(x0 -l a1) ⊦ a = a0, dups(x0), a0 ∈ x0, a0 ∈ x0 -l a

Rule argument

Additional information

Used simplifier rules: del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);%in-01: ⊦ a ∈ x -l b → a ∈ x ∧ a ≠ b;%in: ⊦ a ∈ x ∧ a ≠ b → a ∈ x -l b;% ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list