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Proof step 11 from proof append-mod (previous step)

(x2 + x0)[n0 + # x2, a] = x2 + x1, n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ x0 = [] ⊦ x0[n0, a] = x1, x0 = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0, x0, a, x1

Additional information

Used terms: n0, x0, a, x1