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Proof step 3 from proof app-03 (previous step)

∀ x, z. x + y0 = z + y0 → x = z, x + (a + y0) = z + (a + y0) ⊦ x = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x + a ' + y0 = z + a ' + y0, ∀ x, z. x + y0 = z + y0 → x = z ⊦ x = z

Rule argument

Additional information

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