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Proof step 17 from proof sublist-append-04 (previous step)

m0 + 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(m0, x0) + firstn(n1, restn(m0, x0)) = firstn(m0 + n1, x0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: 0, m0, n1 abort: F discard: F computed: F

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: 0, m0, n1