FDL > PVS > Number Theory > gcd > gcd def > pf:gcd def > 1 > 2 > 1 > 1 > 1 > 1 (4 nodes)

Conclusion

-1. (max({k:posnat | divides(k, i!1) AND divides(k, j!1)}) = nn!1) IFF maximum?(nn!1, {k:posnat | divides(k, i!1) AND divides(k, j!1)})
-2. divides(nn!1, i!1)
-3. divides(nn!1, j!1)
-4. FORALL mm : (divides(mm, i!1) AND divides(mm, j!1)) IMPLIES (mm < = nn!1)
-5. (i!1 /= 0) OR (j!1 /= 0)

1. max({k:posnat | divides(k, i!1) AND divides(k, j!1)}) = nn!1