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Proof step 14 from proof eq-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ¬ i < j ↔ j ≤ i
Proof step 16:
⊦ ¬ i ≤ 0 ∧ ¬ 0 ≤ j, ¬ 0 ≤ i ∧ ¬ j ≤ 0, i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j

Rule argument

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

Additional information

Used terms: j, 0