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Proof step 6 from proof congmod-def-mod-02 (previous step)

0 < abs(j), i < 0, i0 < 0 ⊦ i |%| abs(j) = i0 |%| abs(j) ↔ i |%| j = i0 |%| j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0, i0 < 0 ⊦ i |%| abs(j) = i0 |%| abs(j) ↔ i |%| j = i0 |%| j, j = 0

Rule argument

Additional information

Used simplifier rules: abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs