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Proof step 3 from proof gcd-le (previous step)

⊦ m = 0, n = 0, gcd(m, n) ≤ m

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ m ≠ 0 ∧ n ≠ 0 → gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

Rule argument

spec: <toplevel> name: gcd-le-both seq: ⊦ m ≠ 0 ∧ n ≠ 0 → gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n subst: [m, n] → [m, n]

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) Used terms: m, n