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Proof step 25 from proof put-firstn-restn (previous step)

#(y0.butlast + y0.last ') = m + 1, y0 = y0.butlast + y0.last ', # y0.butlast ≤ m, m ≤ # y0.butlast ⊦ y0.butlast + y0.last ' = [], y0.butlast + y0.last '[m - # y0.butlast, a] = firstn(m, y0.butlast) + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# y0 = m + 1 ⊦ y0.butlast = firstn(m, y0.butlast), y0 = []

Rule argument

Additional information

Used simplifier rules: put-one: ⊦ a '[0, b] = b ';% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ x + y = z + y ↔ x = z; (s)% ⊦ m - m = 0; (s)% ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ x ≠ [] → # x.butlast = # x - 1; (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)% ⊦ # [] = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%x ≠ [] ⊦ x.butlast + x.last ' = x; (s)