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Proof step 1 from proof longest-common-prefix (conclusion)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ ∃ z, x0, y0. x = z + x0 ∧ [] = z + y0 ∧ (x0 ≠ [] ∧ y0 ≠ [] → x0.head ≠ y0.head)
Proof step 3:
∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.head ≠ y1.head) ⊦ ∃ z, x0, y1. x = z + x0 ∧ a + y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.head ≠ y1.head)

Rule argument

term: y

Additional information

Used simplifier rules: list freely generated by [], +;;