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Proof step 16 from proof mod-ex-div (previous step)

j1 < 0, i0 * j + j1 < 0, j < 0, 0 ≤ i0, j < j1 ⊦ j = 0, ∃ j0. i0 * j = j * j0 + j

User applied rule exists right

And obtained the following new goals

Proof step 17:
j1 < 0, i0 * j + j1 < 0, j < 0, 0 ≤ i0, j < j1 ⊦ j = 0, i0 * j = j * (i0 - 1) + j

Rule argument

pos: right 2 terms: i0 - 1 abort: F discard: T computed: F

Additional information

Used terms: i0 - 1