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Proof step 4 from proof between-is-aligned (previous step)

m1 % m = 0, m0 % m = 0, m0 < m1, m1 < m0 + m ⊦ m = 0, m = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)

Rule argument

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

Additional information

Used simplifier rules: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → m % n = m; (s) Used terms: m1, m0