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Proof step 26 from proof cong-equiv (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 27:
n0 = n1 + n, n ≤ n0 ⊦ m | n1 ↔ n % m = n0 % m, m = 0

Rule argument

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

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n, n1