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Proof step 6 from proof div-rec-01 (previous step)

i / j = 1 - (j - i) / j, 1 = i / j + (j - i) / j, j < 0, i < 0, i ≤ j ⊦ i / j = (i - j) / j + 1

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: minus-add: j0 ≠ 0 ⊦ (i - j) / j0 + (j - i) / j0 = 0;% ⊦ i + (j - i) = j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)