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Proof step 1 from proof split-rest (conclusion)

n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # z = n

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # y = n
Proof step 3:
n ≤ # x, ∃ y, z. x = y + z ∧ # y = # x - n ⊦ ∃ y, z. x = y + z ∧ # z = n

Rule argument

spec: <toplevel> name: split-first seq: n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # y = n subst: [n, x] → [# x - n, x]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: # x - n, x