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Proof step 4 from proof gcd-le-both (previous step)

m < n ⊦ m = 0, n = 0, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
m < n ⊦ n = 0, m = 0, ¬ m < gcd(m, n) ∧ ¬ n < gcd(m, n)

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss) Used terms: m, n