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Proof step 4 from proof mod-ls-same-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i ≤ j ↔ ¬ j < i
Proof step 6:
i < 0 ⊦ j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ ~ i < abs(j)

Rule argument

spec: int-abs inst: name: le-def seq: ⊦ i ≤ j ↔ ¬ j < i subst: [j, i] → [~ i, abs(j)] flags: rewrite

Additional information

Used terms: ~ i, abs(j)