Solitaire Language. A 1 B 011 C 110 D 11(01)^{∗}00(10)^{∗}11 E 11(01)^{∗}00(11)^{+}(10)^{∗}11 F 11(01)^{∗}(11)^{+}00(10)^{∗}11 G 11(01)^{∗}(11)^{∗}1011(10)^{∗}11 H 11(01)^{∗}1101(11)^{∗}(10)^{∗}11 I 11(01)^{∗}(11)^{∗}01 J 10(11)^{∗}(10)^{∗}11 Hopping: 1101 → 0011 and so 11(01)^{∗} → 11. Similarly, (10)^{∗}11 → 11. So we can reduce the above cases to A 1 B 011 C 110 D 110011 E 1100(11)^{+} F (11)^{+}0011 G (11)^{+}101111 H 111101(11)^{∗} I 110 J 011 Then

• B → A

• C → A

• D → I

• E → J

• F → I

• G → F

• H → E Unhopping 001 → 110 or 100 → 011. So,

• A ← B ∨ C

• B ← I ∨ J

• C ← I ∨ J

• D ← D ∨ E ∨ F • E ← E ∨ H

• F ← F ∨ G

• G ← G

• H ← H

• I ← I ∨ D ∨ F

• J ← J ∨ E ∨ D