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Proof step 24 from proof eq-one (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 25:
i * j = 1, 0 ≤ j, 0 < i, j ≤ 1, 0 ≤ i ⊦ i = 1

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs