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Proof step 1 from proof mod-eq-02 (conclusion)
m0 < n, n ≠ 0 ⊦ (m + (* m0)) % (* n) = m0 ↔ m % n = 0
Special step normalize
And obtained the following new goals
Proof step 2
:
m0 < n, n ≠ 0 ⊦ (m + m0) % n = m0 ↔ m % n = 0