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Proof step 7 from proof le-div-to-mult (previous step)

n * m0 ≤ m1 * m0 + n0, n0 < m0, m1 < n ⊦ m0 = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 9:
n * m0 ≤ m1 * m0 + n0, n0 < m0, ∃ m0. m1 + m0 = n ∧ m0 ≠ 0 ⊦ m0 = 0, n = 0

Rule argument

spec: nat-basic2 inst: name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [n, m1] flags: rewrite

Additional information

Used terms: n, m1