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Proof step 7 from proof prefixn-butlast (previous step)

# y = m + 1, x = y + z, m < # x, m + 1 ≤ # x ⊦ y.butlast = firstn(m, x), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# y = m + 1 ⊦ y + z = [], y.butlast = firstn(m, y + z)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)