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Proof step 21 from proof gcd-divides-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
j | abs(i0) ⊦ j | i0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ abs(i) < 0; (s)