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Proof step 6 from proof div-ngr-zero (previous step)

abs(j0) < j, 0 < j, i0 < j, j0 < j, i0 * j + j0 < 0, i0 ≤ 0, j0 ≤ 0 ⊦ i0 = 0 ↔ 0 < j + i0 * j + j0

User applied rule cut formula

And closed the goal!

Rule argument

fma: i0 = 0

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 ∧ j0 < j ∧ 0 < j → i * j + j0 < 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ 0 < j → (0 < i * j + j ↔ 0 ≤ i); (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i * 0 = 0; (s)% ⊦ i + 0 = i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s) Used formulas: i0 = 0