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Proof step 10 from proof le-div-to-mult (previous step)

(m1 + m) * m0 ≤ m1 * m0 + n0, n0 < m0 ⊦ m0 = 0, m1 + m = 0, m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 12:
(m1 + m) * m0 ≤ m1 * m0 + n0, ∃ m1. n0 + m1 = m0 ∧ m1 ≠ 0 ⊦ m0 = 0, m1 + m = 0, m = 0

Rule argument

spec: nat-basic2 inst: name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [m0, n0] flags: rewrite

Additional information

Used terms: m0, n0