FDL > PVS > Number Theory > gcd > gcd def > pf:gcd def > 1 > 1 (7 nodes)

Conclusion

-1. (i!1 /= 0) OR (j!1 /= 0)

1. (gcd(i!1, j!1) = nn!1) IMPLIES ((divides(nn!1, i!1) AND divides(nn!1, j!1)) AND (FORALL mm : (divides(mm, i!1) AND divides(mm, j!1)) IMPLIES (mm < = nn!1)))

Tactic
 FLATTEN
Premise 1.   (has proof of 6 steps)

-1. gcd(i!1, j!1) = nn!1
-2. (i!1 /= 0) OR (j!1 /= 0)

1. (divides(nn!1, i!1) AND divides(nn!1, j!1)) AND (FORALL mm : (divides(mm, i!1) AND divides(mm, j!1)) IMPLIES (mm < = nn!1))