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Proof step 11 from proof lastn-restn (previous step)

m < #(a ' + x0), m + m0 ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, (a ' + x0).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m + m0 ≤ # x0, m ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, x0)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list