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Proof step 1 from proof elim-first (conclusion)

⊦ x = [] ↔ ¬(∃ a, y. x = a ' + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ x = [], ∃ a, y. x = a ' + y

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ [];%a: ⊦ (x + y) + z = x + y + z;