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Proof step 5 from proof div-div-01 (previous step)

j | i, 0 ≤ i, 0 ≤ j, ∃ i0. i * i0 = j ⊦ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, i * i0 | i, 0 ≤ i0 ⊦ i = 0, i0 = 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j = i ↔ i = 0 ∨ j = 1; (ws)