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Proof step 16 from proof gcd-divides-both (
previous step
)
i = ~ abs(i0), i < 0, j | i ⊦ i0 = 0, j | i0
User applied rule insert equation
And obtained the following new goals
Proof step 17
:
~ abs(i0) < 0, j | ~ abs(i0) ⊦ i0 = 0, j | i0
Rule argument
pos: left 1 flags: discard
Additional information