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Proof step 7 from proof divides-abs-02 (previous step)

i * i0 = abs(j) * j0 ⊦ ∃ i0. i * i0 = j * j0

User applied rule exists right

And obtained the following new goals

Proof step 8:
i * i0 = abs(j) * j0 ⊦ i * i0 = j * j0, ∃ i0. i * i0 = j * j0

Rule argument

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

Additional information

Used terms: i0