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Proof step 4 from proof mod-twice (previous step)

m ≤ n0, ∀ n, m, n1. n < n0 → (true → m = 0 ∨ (n % m + n1) % m = (n + n1) % m) ⊦ m = 0, ((n0 - m) % m + n1) % m = (n0 + n1) % m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n0 = n2 + m, m ≤ n0, ∀ n, m, n1. n < n0 → (true → m = 0 ∨ (n % m + n1) % m = (n + n1) % m) ⊦ m = 0, (n2 % m + n1) % m = (n0 + n1) % m

Rule argument

[x0lemmaarg: spec-name: nat instance-name: lemma-name: elim-sub antecedent: n ≤ m succedent: n0 = m - n ↔ m = n0 + n substlist: [m, n, n0] → [n0, m, n2] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≤ m ⊦ n0 = m - n ↔ m = n0 + n])], [EXTRATERMS : (n0, m, n2)]]