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Proof step 11 from proof longest-common-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
x ≠ [] ∧ y ≠ [] → x.head ≠ y.head ⊦ ∃ z0, x0, y1. a0 ' + z + x = z0 + x0 ∧ a0 ' + z + y = z0 + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.head ≠ y1.head)

Proof step 11 from proof longest-common-prefix (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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