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Proof step 1 from proof one-01 (conclusion)

⊦ y ≠ [] → (a ' = x + y ↔ x = [] ∧ a ' = y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-04: ⊦ a ' = b ' ↔ a = b;