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Proof step 19 from proof mod-ls-same-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ i ≤ j ↔ ¬ j < i
Proof step 21:
0 < i ⊦ abs(j) = 1, j = 1, j = 0, i = 0, i % j < i ↔ ¬ 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)