Proofs in Polyadic QL - Solutions

3. (x)(y)(Rxy ⊃ ~Ryx) /// (x)~Rxx

Proof:

1. (x)(y)(Rxy ⊃ ~Ryx) premise  
2. flag a flagging assump., U.G. sp1
3. (y)(Ray ⊃ ~Rya) U.I., 1 sp1
4. (Raa ⊃ ~Raa) U.I. 3 sp1
5. ~Raa v ~Raa C.E., 4 sp1
6. ~Raa rep, 5 sp1
7. (x)~Rxx U.G. 2-6