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Proof step 6 from proof lastN-append-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # y, n ≤ # x0 + # y, ∀ n, y. n ≤ # y → lastn(n, x0 + y) = lastn(n, y) ⊦ lastn(n, x0 + y) = lastn(n, y)

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: ⊦ #(x + y) = # x + # y; from specification list%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