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Proof step 3 from proof congmod-mod-abs (previous step)

⊦ j | abs(i - i0) ↔ ≡mod(i, i0, abs(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ j | i - i0 ↔ ≡mod(i, i0, abs(j))

Rule argument

Additional information

Used simplifier rules: div-abs-02: ⊦ i | abs(j) ↔ i | j; from specification int-gcd