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Proof step 9 from proof blastN-firstN (previous step)

1 + # x0 - n ≠ 0, n ≤ # x0 + 1, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = a ' + firstn((1 + # x0 - n)-1, x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n ≤ # x0, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ 1 + # x0 = n, butlastn(n, a ' + x0) = a ' + firstn(# x0 - n, x0)

Rule argument

Additional information

Used simplifier rules: add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%zero-02: ⊦ m ≤ n → (n - m = 0 ↔ n = m); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%pred-03: ⊦ n -1 = n - 1; from specification nat