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Proof step 18 from proof mod-neq (previous step)

i * j = i * j0 + j1, abs(j1) < i, j1 < i, 0 ≤ i ⊦ j1 = 0, 0 < j1 ∧ 0 < i, 0 < i ∧ j1 < 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s)% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ i < j → ¬ j < i; (s)