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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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: ⊦ 0 + m = m; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ n → (n - m = 0 ↔ n = m); (s)% ⊦ n -1 = n - 1; (s)