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Proof step 14 from proof rec (previous step)

⊦ dups(a ' + x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z)))), a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ dups(a ' + x) ↔ (∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z), a ∈ x

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list