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Proof step 7 from proof divides-abs-01 (
previous step
)
⊦ ∃ i1. abs(i) * i1 = i * i0
User applied rule exists right
And obtained the following new goals
Proof step 8
:
⊦ abs(i) * i0 = i * i0, ∃ i1. abs(i) * i1 = i * i0
Rule argument
pos: right 1 terms: i0 abort: F discard: F computed: F
Additional information
Used terms: i0