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Proof step 5 from proof prefixn-butlast (previous step)

n = m + 1, n ≤ # x ⊦ firstn(n, x).butlast = firstn(m, x), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m < # x ⊦ firstn(m + 1, x).butlast = firstn(m, x), x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)