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Proof step 3 from proof app-04 (previous step)

∀ y. x0 = x0 + y → y = [], a + x0 = (a + x0) + y ⊦ y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x0 = x0 + y, ∀ y. x0 = x0 + y → y = [] ⊦ y = []

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%app-08: ⊦ a ' + x = a ' + y ↔ x = y;