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Proof step 2 from proof eq-one (previous step)

⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
0 < i, 0 < j, i ≤ i0, j < j0 ⊦ i * j < i0 * j0
Proof step 4:
⊦ 0 < 1 ∧ 0 < 1 ∧ 1 ≤ i ∧ 1 < j, i * j = 1 ↔ abs(j) = 1 ∧ i = j
Proof step 29:
1 * 1 < i * j ⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j

Rule argument

spec: int-mult inst: name: pls-pos seq: 0 < i, 0 < j, i ≤ i0, j < j0 ⊦ i * j < i0 * j0 subst: [i, j, i0, j0] → [1, 1, i, j] flags: current, []

Additional information

Used terms: 1, 1, i, j