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Proof step 10 from proof gcd-le-both (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 11:
m < n ⊦ n = 0, m = 0, ¬ m < gcd(m, n) ∧ ¬ n < gcd(m, n)

Rule argument

spec: nat-prime inst: name: gcd-le-both-h seq: m < n ⊦ n = 0, m = 0, ¬ m < gcd(m, n) ∧ ¬ n < gcd(m, n) subst: [m, n] → [n, m] flags: current, []

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat Used terms: n, m