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Proof step 6 from proof dups (previous step)

dups((a0 ' + x0) -1l a), ∀ a1. ¬ dups(x0 -1l a1) ⊦ a = a0, dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%del1-in: ⊦ a ≠ b → (a ∈ x -1l b ↔ a ∈ x);% ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list