Stefan Buecker wrote on 10/30/11 at 12:44:34:
In my opinion 1.e4 e5 2.Nf3 Nc6 3.Bb5 f5 4.Nc3 fxe4 5.Nxe4 Nf6 6.Nxf6+ Qxf6 7.Qe2 does not work, because 7...Be7 8.Bxc6 bxc6 is =.
I hope you will forgive my scepticism, but I need a little more to be convinced. OK, it wouldn't surprise me if accepting the pawn provides compensation (although this needs to be proven yet) but I quite like 9.d4. Three sample lines:
a) 9...exd4 10.Bg5 Qe6 11.Qxe6 dxe6 12.Bxe7 Kxe7 13.Nxd4 and White has a very nice and stable advantage, Janovsky-Tseshkovsky, Voskresensk 1992.
b) 9...O-O 10.Bg5 Qe6 (Qg6 11.Bxe7+ Re8 12.O-O Rxe7 13.Nxe5 is excellent for White as the pin always can be be resolved by Qc4+) 11.Bxe7 Qxe5 12.dxe5 and White is at least a bit better, Adorjan-Cortlever, Amsterdam 1970.
c) 9...Qg6 10.O-O d6 11.Qc4 d5 12.Qc3 exd4 13.Nxd4
Bd7 14.Re1 Qf6? (better Kf7 but 15.Bf4 remains nice for White; the Knight is not weaker than the Bishops, but Black's pawns are weaker than White's) 15.Nxc6 Qxc6 16.Qxg7 and White has the attack, Copar-Auzins, corr ICCF 1997.
13...c5!! =.