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Proof step 20 from proof gcd-divides-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: absls: ⊦ ¬ abs(i) < 0; from specification int-abs