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Proof step 11 from proof modex (previous step)

i % j = 0 ⊦ j = 0, ∃ j0. i = j0 * j, 0 < j ∧ 0 ≤ i, j < 0 ∧ 0 ≤ i, 0 < j ∧ i < 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ i ≤ j ↔ ¬ j < i
Proof step 13:
i % j = 0 ⊦ j = 0, ∃ j0. i = j0 * j, 0 < j ∧ i < 0, j < 0 ∧ ¬ i < 0, 0 < j ∧ ¬ i < 0

Rule argument

spec: int-abs name: le-def seq: ⊦ i ≤ j ↔ ¬ j < i subst: [j, i] → [i, 0] flags: <all paths>

Additional information

Used terms: i, 0