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Proof step 3 from proof mod-neq-03 (previous step)

m0 * n + n0 + 1 = m * n + n1, n1 < n, n1 < n0, n0 < n, n1 + 1 < n ⊦ n1 = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1)
Proof step 5:
m0 * n + n0 + 1 = m * n + n1, n1 + 1 < n, n0 < n, n1 < n0, n1 < n, n0 + 1 < n ∧ n1 < n → (m0 * n + n0 + 1 = m * n + n1 ↔ m0 = m ∧ n0 + 1 = n1) ⊦ n1 = 0

Rule argument

spec: <toplevel> name: modeq seq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1) subst: [n0, n, n1, m, m0] → [n0 + 1, n, n1, m0, m]

Additional information

Used terms: n0 + 1, n, n1, m0, m