3 ms·
When !E(y,z), then also !E(z,y) (because symmetry). Also, E(y,x). Thus !E(x,z), because E(y,x) and E(x,z) would imply E(y,z), which is false.
by sauerbraten 13y ago
When !E(y,z), then also !E(z,y) (because symmetry).
Also, E(y,x).
Thus !E(x,z), because E(y,x) and E(x,z) would imply E(y,z), which is false.