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Proof step 21 from proof sublist-append-04 (previous step)

m + n0 + n1 ≤ # x0, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(m, x0)) + firstn(n1, restn(m + n0, x0)) = firstn(n0 + n1, restn(m, x0))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m, n0, n1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m, n0, n1