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Proof step 4 from proof del-del1-nodups (previous step)

∀ a. x0 -l a = x0 -1l a ∨ dups(x0) ⊦ (a0 + x0) -l a = (a0 + x0) -1l a, dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a1. x0 -l a1 = x0 -1l a1 ⊦ (a0 ' + x0) -l a = (a0 ' + x0) -1l a, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ a + x = a ' + x; (s)