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Proof step 7 from proof longest-common-prefix (previous step)

x = a0 + x1, ∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first) ⊦ ∃ z, x0, y. x = z + x0 ∧ a ' + y0 = z + y ∧ (x0 ≠ [] ∧ y ≠ [] → x0.first ≠ y.first)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first) ⊦ ∃ z, x, y. a0 ' + x1 = z + x ∧ a ' + y0 = z + y ∧ (x ≠ [] ∧ y ≠ [] → x.first ≠ y.first)

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;