YES

We show the termination of the TRS R:

  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)))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)
p2: h#(c(s(x),c(s(|0|()),y)),z) -> h#(y,c(s(|0|()),c(x,z)))

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)
p2: h#(c(s(x),c(s(|0|()),y)),z) -> h#(y,c(s(|0|()),c(x,z)))

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,3)
            c_A(x1,x2) = x1 + x2
            s_A(x1) = ((0,0),(1,0)) x1
            |0|_A() = (2,1)
    
      precedence:
      
        h# = c = s = |0|
      
      partial status:
    
        pi(h#) = []
        pi(c) = []
        pi(s) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1,x2) = (0,0)
            c_A(x1,x2) = x2 + (3,1)
            s_A(x1) = (2,2)
            |0|_A() = (1,0)
    
      precedence:
      
        |0| > h# = c = s
      
      partial status:
    
        pi(h#) = []
        pi(c) = [2]
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1,x2) = ((1,0),(0,0)) x2 + (3,1)
            c_A(x1,x2) = x2 + (2,2)
            s_A(x1) = (1,0)
    
      precedence:
      
        c > s > h#
      
      partial status:
    
        pi(h#) = []
        pi(c) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1,x2) = (0,0)
            c_A(x1,x2) = (0,0)
            s_A(x1) = (0,0)
    
      precedence:
      
        c = s > h#
      
      partial status:
    
        pi(h#) = []
        pi(c) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.