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Proof step 1 from proof eq-04 (conclusion)

⊦ x + a ' + y = b ' ↔ x = [] ∧ a = b ∧ y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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