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Proof step 2 from proof longest-common-prefix (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], x, [] abort: F discard: T computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z; Used terms: [], x, []