YES

Problem:
 h(x,c(y,z)) -> h(c(s(y),x),z)
 h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))

Proof:
 DP Processor:
  DPs:
   h#(x,c(y,z)) -> h#(c(s(y),x),z)
   h#(c(s(x),c(s(0()),y)),z) -> h#(y,c(s(0()),c(x,z)))
  TRS:
   h(x,c(y,z)) -> h(c(s(y),x),z)
   h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))
  Bounds Processor:
   bound: 1
   enrichment: match-dp
   automaton:
    final states: {1}
    transitions:
     c0(17,18) -> 18*
     c0(2,18) -> 15*
     c0(17,15) -> 18*
     c0(3,2) -> 4*
     c0(3,4) -> 4*
     c0(3,22) -> 4*
     c0(2,2) -> 15*
     s0(2) -> 3*
     s0(16) -> 17*
     00() -> 16*
     h{#,1}(22,18) -> 1*
     h{#,1}(20,15) -> 1*
     h{#,1}(20,18) -> 1*
     h{#,1}(22,2) -> 1*
     c1(19,2) -> 20*
     c1(19,4) -> 20*
     c1(19,20) -> 20*
     c1(19,22) -> 20*
     c1(21,20) -> 22*
     s1(17) -> 19*
     s1(2) -> 21*
     f50() -> 2*
     h{#,0}(2,18) -> 1*
     h{#,0}(4,2) -> 1*
     h{#,0}(4,18) -> 1*
   problem:
    DPs:
     h#(c(s(x),c(s(0()),y)),z) -> h#(y,c(s(0()),c(x,z)))
    TRS:
     h(x,c(y,z)) -> h(c(s(y),x),z)
     h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))
   Subterm Criterion Processor:
    simple projection:
     pi(h#) = 0
    problem:
     DPs:
      
     TRS:
      h(x,c(y,z)) -> h(c(s(y),x),z)
      h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))
    Qed