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Proof step 1 from proof prefix-app-03 (conclusion)

⊦ x + y + z ⊑ y ↔ x = [] ∧ z = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + y + z ⊑ y ⊦ x = [] ∧ z = []

Rule argument

Additional information

Used simplifier rules: prefix-self: ⊦ x ⊑ x;%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)% ⊦ [] + x = x; (s)