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Proof step 11 from proof divides-add (
previous step
)
i * i1 = i * i0 + j0 ⊦ ∃ i0. i * i0 = j0
User applied rule exists right
And obtained the following new goals
Proof step 12
:
i * i1 = i * i0 + j0 ⊦ i * (i1 - i0) = j0
Rule argument
pos: right 1 terms: i1 - i0 abort: F discard: T computed: F
Additional information
Used terms: i1 - i0