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Proof step 8 from proof three (previous step)

i * j = 3, 0 < i, 1 < i, j < i, j ≤ 1 ⊦ i = 3 ∧ j = 1, 3 ≤ i ∧ 1 < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ 0 < i → (i ≤ 1 ↔ i = 1)
Proof step 10:
i * j = 3, j < i, 1 < i, 0 < i, j ≤ 1 ⊦ 0 < j, 3 ≤ i ∧ 1 < j, i = 3 ∧ j = 1

Rule argument

spec: int-abs name: bet seq: ⊦ 0 < i → (i ≤ 1 ↔ i = 1) subst: [i] → [j] flags: <all paths>

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i; Used terms: j