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Proof step 10 from proof cong-mult (previous step)

j0 * j + i = i1 + i0, j ≤ 0 ⊦ j | i1 ↔ j | abs(i - i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j0 * j + i = i1 + i0, j ≤ 0 ⊦ j | i1 ↔ j | i - i0

Rule argument

Additional information

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