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Proof step 6 from proof ls-01 (previous step)

m1 * n0 + n1 < m1 * n, m < m1 * n, n1 < n0, m1 * n0 + n1 ≤ m, n ≤ n0 ⊦ m1 = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 8:
n1 < n0, m < m1 * n, m1 * n0 + n1 < m1 * n, m1 * n0 + n1 ≤ m, ∃ n1. n + n1 = n0 ⊦ m1 = 0, n = 0

Rule argument

spec: nat name: le-addex seq: ⊦ m ≤ n ↔ (∃ n0. m + n0 = n) subst: [n, m] → [n0, n] flags: <all paths>

Additional information

Used terms: n0, n