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Proof step 4 from proof longest-common-prefix (previous step)

∀ 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)

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
x = [], ∀ 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)
Proof step 7:
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)

Rule argument

var: x

Additional information

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