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Proof step 9 from proof eq-one (previous step)

⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j, i < 0 ∧ 0 < j, ¬ i ≤ 0 ∧ j < 0, ¬ i ≤ 0 ∧ 1 < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i < j ↔ ¬ j ≤ i
Proof step 11:
⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j, ¬ i ≤ 0 ∧ ¬ 0 ≤ j, i < 0 ∧ 0 < j

Rule argument

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

Additional information

Used terms: 0, j