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Proof step 6 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) = firstn(# x0 + 1 - n, a ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(# x0, (a ' + x0).butlast) = []

Rule argument

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%ler: ⊦ m ≤ m; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%zero: ⊦ m - m = 0; from specification nat%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast); from specification list-last%zero-02: ⊦ m ≤ n → (n - m = 0 ↔ n = m); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2