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Proof step 6 from proof lastn-restn (previous step)

0 < n, m + n ≤ # x0 + 1, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(n -1, (a ' + x0).rest)) = lastn(m, a ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + n ≤ # x0 + 1, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(n - 1, x0)) = lastn(m, a ' + x0), n = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat