[Imprint]
[Data Privacy Statement]
[E-Mail]
View
long theorem info
. View
the proof
. Back to
main index
. View
the specification
. View
theorem base summary
, View
theorem base listing
, the
used properties by spec
, the
dependency list
, the
symbol table
,
Proof step 3 from proof gcd-div-gcd (
previous step
)
gcd(i, j) = gcd(abs(i), abs(j)) ⊦ i ≠ 0 → gcd(i / gcd(i, j), j / gcd(i, j)) = 1
User applied rule insert equation
And obtained the following new goals
Proof step 4
:
gcd(i, j) = gcd(abs(i), abs(j)) ⊦ i ≠ 0 → gcd(i / gcd(abs(i), abs(j)), j / gcd(abs(i), abs(j))) = 1
Rule argument
pos: left 1 flags:
Additional information