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Proof step 18 from proof put-firstn-restn (previous step)

# y0 = m + 1, m < # y0 ⊦ y0[m, a] + z = firstn(m, y0) + a ' + z, y0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# y0 = m + 1 ⊦ y0[m, a] = firstn(m, y0) + a ', y0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = z + y ↔ x = z; (s)