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Proof step 22 from proof div-eq-same-01 (previous step)

0 = i0 * i + j0, j0 < 0, 0 < i0, j0 < i0, i < 0, 0 ≤ j0 + i0, j0 ≤ i0 ⊦ ~1 = i0, 0 = i0, j0 = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: 1 * i + j0 < i0 * i

Additional information

Used simplifier rules: ⊦ 1 * i = i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i < j ↔ ¬ j ≤ i; (ss)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1% ⊦ i +1 = i + 1; (s)% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: 1 * i + j0 < i0 * i