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Proof step 12 from proof gcd-divides-both (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ ¬ j < 0 → abs(j) = j
Proof step 14:
abs(i0) = abs(i), i < 0, j | abs(i) ⊦ j | i ∧ j | i0
Proof step 21:
abs(i0) = i, j | i ⊦ i < 0, j | i ∧ j | i0

Rule argument

spec: int-abs name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [i] flags: <all paths>

Additional information

Used terms: i