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Proof step 5 from proof eq-one (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
0 < i ∧ j < 0 ∨ i < 0 ∧ 0 < j ⊦ i * j < 0
Proof step 7:
⊦ 0 < i ∧ j < 0 ∨ i < 0 ∧ 0 < j, i * j = 1 ↔ abs(j) = 1 ∧ i = j, 0 < i ∧ 1 < j
Proof step 28:
i * j < 0 ⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j, 0 < i ∧ 1 < j

Rule argument

spec: int-mult inst: name: pzero-ls seq: 0 < i ∧ j < 0 ∨ i < 0 ∧ 0 < j ⊦ i * j < 0 subst: [i, j] → [i, j] flags: current, []

Additional information

Used terms: i, j