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Proof step 22 from proof sublist-move (previous step)

m1 ≤ # x0, m2 + m ≤ m1, m2 + m ≤ # x0, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(m2, firstn(m1, x0))) = firstn(m, restn(m2, x0))

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m2, m, m1, 0 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: m2, m, m1, 0