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Proof step 7 from proof cong-equiv (previous step)

n0 < n, n0 < n ⊦ m = 0, m | n - n0 ↔ n % m = n0 % m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
n = n1 + n0 ∧ n0 ≤ n, n0 < n, n0 < n ⊦ m = 0, m | n1 ↔ n % m = n0 % m

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n1] flags:

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n1