Easy Proofs in QL

Prove the following arguments. Click on selected problems for solutions.

1. (x)Sx ⊃ (∃x)Px, (x)~Px & (∃x)Gx  /// (∃x)~Sx
   
2. (x)Fx /// (∃x)Fx
   
3. (x)Ax ⊃ (∃x)Bx, (∃x)Bx ⊃ ~(x)Cx, /// (x)Ax ⊃ (∃x)~Cx
   
4. (∃x)[(Ax ⊃ (Bx v ~Cx)], (∃x)Cx ⊃ (x)Ax, (x)~Bx & (x)Ax  ///~(x)Cx
   
5. (x)(Rx v ~Sx), (x)(~Sx ⊃ Tx), ~Ta ///  (∃x)Rx
   
6. Ta & ~Sa, (x)(~Sx ⊃ (∃y)Ry) /// (∃x)(Rx v Mx)
   
7. (∃x)Fx /// (x)Gx  ⊃ (∃x)(Fx & Gx)
   
8. (∃x)(Ax & Bx), (x)(Bx ⊃ ~Cx) // (∃x)(~Cx & Ax)
   
9. (x)(Sx ⊃ Tx) // (∃x)(~Sx v Tx)
   
10. (x)Ax ⊃ (x)Bx, ~(∃x)~Ax // Ba
    
11. (x)(~Ax v Bx), (∃x)Ax // (∃x)(Ax & Bx)
    
12. ~(∃x)(Ax & ~Bx), ~(∃x)(Bx & ~Cx) // (x)(Ax ⊃ Cx)
    
13. (x)~(~Ax v ~Bx), ~(x)Bx v Da // (∃x)Dx
    
14. (x)(Tx ⊃ (Rx v Px), (∃x)(Tx & Lx), (x)(Rx ⊃ ~Tx) // (∃x)Px
    
15. (x)(Sx ⊃ Tx) // (∃x)~Sx v (∃x)Tx  [note: this is a variation on 9, but a bit harder - hint: use indirect proof]