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

x + a ' + (a0 + y0) = b ', ∀ x, a, b. x + a ' + y0 = b ' → x = [] ∧ a = b ∧ y0 = [] ⊦ x = [] ∧ a = b ∧ a0 + y0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x + a ' + a0 ' + y0 = b ', ∀ x, a, b. x + a ' + y0 = b ' → x = [] ∧ a = b ∧ y0 = [] ⊦

Rule argument

Additional information

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