YES

Problem:
 q(0(x1)) -> p(p(s(s(0(s(s(s(s(x1)))))))))
 q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
 r(0(x1)) -> p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
 r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
 p(p(s(x1))) -> p(x1)
 p(s(x1)) -> x1
 p(0(x1)) -> 0(s(s(s(x1))))

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [r](x0) = 12x0,
   
   [p](x0) = x0,
   
   [s](x0) = x0,
   
   [q](x0) = 12x0,
   
   [0](x0) = x0
  orientation:
   q(0(x1)) = 12x1 >= x1 = p(p(s(s(0(s(s(s(s(x1)))))))))
   
   q(s(x1)) = 12x1 >= 12x1 = p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
   
   r(0(x1)) = 12x1 >= x1 = p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
   
   r(s(x1)) = 12x1 >= 12x1 = p(s(p(s(s(q(p(s(p(s(x1))))))))))
   
   p(p(s(x1))) = x1 >= x1 = p(x1)
   
   p(s(x1)) = x1 >= x1 = x1
   
   p(0(x1)) = x1 >= x1 = 0(s(s(s(x1))))
  problem:
   q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
   r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1))))))))))
   p(p(s(x1))) -> p(x1)
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(s(x1))))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {25,2,24,15,1}
    transitions:
     p1(27) -> 28*
     p1(33) -> 34*
     f50() -> 2*
     p0(5) -> 6*
     p0(17) -> 18*
     p0(2) -> 24*
     p0(14) -> 1*
     p0(4) -> 5*
     p0(21) -> 22*
     p0(23) -> 15*
     p0(13) -> 14*
     p0(3) -> 16*
     s0(20) -> 21*
     s0(10) -> 11*
     s0(22) -> 23*
     s0(12) -> 13*
     s0(7) -> 8*
     s0(2) -> 3*
     s0(19) -> 20*
     s0(9) -> 10*
     s0(4) -> 26*
     s0(16) -> 17*
     s0(11) -> 12*
     s0(8) -> 9*
     s0(3) -> 4*
     r0(6) -> 7*
     q0(18) -> 19*
     00(26) -> 25*
     1 -> 19*
     2 -> 24,34,16
     3 -> 5,33
     11 -> 28,1
     12 -> 14,27
     15 -> 7*
     16 -> 18*
     20 -> 22*
     22 -> 15*
     25 -> 24*
     28 -> 1*
     34 -> 6*
   problem:
    
   Qed