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Proof step 10 from proof div-eq-same-01 (previous step)

i / j = i ⊦ j = 0, i = 0, j = 1, j < 0 ∧ i < 0, 0 < j ∧ 0 < i

User applied rule cut formula

And obtained the following new goals

Proof step 11:
0 < j, i / j = i ⊦ j = 0, i = 0, j = 1, j < 0 ∧ i < 0, 0 < j ∧ 0 < i

Rule argument

fma: 0 < j

Additional information

Used simplifier rules: div-zero-ngr-01: ⊦ 0 ≤ i ∧ j < 0 → ¬ 0 < i / j;%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%zero-gels: ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs Used formulas: 0 < j