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Proof step 12 from proof longest-common-prefix (
previous step
)
x ≠ [] ∧ y ≠ [] → x.first ≠ y.first ⊦ ∃ z0, x0, y1. a0 ' + z + x = z0 + x0 ∧ a0 ' + z + y = z0 + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first)
User applied rule exists right
And closed the goal!
Rule argument
pos: right 1 terms: a0 ' + z, x, y abort: F discard: F computed: T
Additional information
Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; Used terms: a0 ' + z, x, y