<div class='user_solution'><div>This study shows a variation of the Prokeš theme names after Ladislav Prokeš: The checkless Prokeš.</div><div><b>1. a7</b> </div><div>1. Kf2? c2 2. a7 cxd1=Q 3. a8=Q Qd2+ {perpetual check=} (3... Qd4+) (3... Qc2+)</div><div><b>1... Kxa7</b> <br />1... c2? 2. a8=Q +- (or 2. Ra1) <br /><b>2. Kf2 <br /></b>2. Ra1+? Kb6 3. Kf2 b2 4. Ra6+ Kxa6= (4... Kb5=) </div><div><b>2... c2</b> <br />2... b2 3. Be4+- <br /><b>3. Ra1+ Kb6<br /></b>3... Kb8 4. Be4 b2 5. Ra8+ (Regular Prokeš theme) Kxa8 6. Bxc2+- </div><div><b>4. Be4 b2 5. Ra7!!</b></div><div>Checkless Prokeš theme. <br />Logical try 5. Ra6+? (Regular Prokeš theme)</div><div>Kxa6 6. Bxc2 Kb5 (the black king is one move closer compared to the main line) 7. Ke3 (7. Be4 Kxb4 8. Ke3) 7... Kxc6 8. Kd4 Kb5 9. Kc3 c5 10.</div><div>Bd3+ Kb6! =) 5... Kxa7 (5... c1=Q 6. Rb7+ Ka6 7. Bd3+ Qc4 8. Bxc4#) (5...</div><div>b1=Q 6. Rb7+ Ka6 7. Bd3#) </div><div><b>6. Bxc2 Kb6 7. Be4!</b> <br />(7. Ke3? Kxc6 8. Kd4 Kb5 9. Kc3 c5 10. Bd3+ Kb6= <b>7... b1=Q <br /></b>(7... Kb5 8. Ke3+- (8. Ke2) (8. Ke1)) <br /><b>8. </b><b>Bxb1 Kxc6 9. Bd3</b> +- 1-0</div><br /></div>