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Proof step 12 from proof prefixn-prefixn (previous step)

# y0 = # y, firstn(m, y0 + z0) = y + z, # y < m, m ≤ # y + # z0 ⊦ y = y0, z0 = []

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-del name: elim-firstn-restn seq: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n) subst: [x, n, y, z] → [y0 + z0, m, y1, z1]

Additional information

Used simplifier rules: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n); (elim) Used terms: y0 + z0, m, y1, z1