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Proof step 6 from proof mod-ls-same-02 (previous step)

i < 0 ⊦ j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ ~ i < abs(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ~ i < j ↔ 0 < j + i
Proof step 8:
i < 0 ⊦ j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i

Rule argument

spec: int-abs inst: name: neg-13 seq: ⊦ ~ i < j ↔ 0 < j + i subst: [j, i] → [abs(j), i] flags: rewrite

Additional information

Used terms: abs(j), i