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Proof step 10 from proof lastN-blastN (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n ≤ # y + 1, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(n, y + a ') + lastn(n - 1, y) = y, n = 0

Rule argument

Additional information

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