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Proof step 6 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. [] = z + x0 ∧ a ' + y0 = z + y ∧ (x0 ≠ [] ∧ y ≠ [] → x0.first ≠ y.first)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], [], a + y0 abort: F discard: T computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c: ⊦ a + x = a ' + x;%app-08: ⊦ a ' + x = a ' + y ↔ x = y; Used terms: [], [], a + y0