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Proof step 13 from proof longest-common-prefix (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: [], a0 ' + x1, a ' + y0 abort: F discard: F computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%app-01: ⊦ x + y = x + z ↔ y = z;%e-01: ⊦ a ' + x ≠ [];%c-08: ⊦ (a ' + x).head = a; Used terms: [], a0 ' + x1, a ' + y0