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Proof step 8 from proof le-div-to-mult (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 10:
n0 < m0, n * m0 ≤ m1 * m0 + n0, ∃ m0. m1 + m0 = n ∧ m0 ≠ 0 ⊦

Rule argument

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

Additional information

Used terms: n, m1