YES

We show the termination of the TRS R:

  h(c(x,y),c(s(z),z),t(w)) -> h(z,c(y,x),t(t(c(x,c(y,t(w))))))
  h(x,c(y,z),t(w)) -> h(c(s(y),x),z,t(c(t(w),w)))
  h(c(s(x),c(s(|0|()),y)),z,t(x)) -> h(y,c(s(|0|()),c(x,z)),t(t(c(x,s(x)))))
  t(t(x)) -> t(c(t(x),x))
  t(x) -> x
  t(x) -> c(|0|(),c(|0|(),c(|0|(),c(|0|(),c(|0|(),x)))))

-- SCC decomposition.

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

p1: h#(c(x,y),c(s(z),z),t(w)) -> h#(z,c(y,x),t(t(c(x,c(y,t(w))))))
p2: h#(c(x,y),c(s(z),z),t(w)) -> t#(t(c(x,c(y,t(w)))))
p3: h#(c(x,y),c(s(z),z),t(w)) -> t#(c(x,c(y,t(w))))
p4: h#(x,c(y,z),t(w)) -> h#(c(s(y),x),z,t(c(t(w),w)))
p5: h#(x,c(y,z),t(w)) -> t#(c(t(w),w))
p6: h#(c(s(x),c(s(|0|()),y)),z,t(x)) -> h#(y,c(s(|0|()),c(x,z)),t(t(c(x,s(x)))))
p7: h#(c(s(x),c(s(|0|()),y)),z,t(x)) -> t#(t(c(x,s(x))))
p8: h#(c(s(x),c(s(|0|()),y)),z,t(x)) -> t#(c(x,s(x)))
p9: t#(t(x)) -> t#(c(t(x),x))

and R consists of:

r1: h(c(x,y),c(s(z),z),t(w)) -> h(z,c(y,x),t(t(c(x,c(y,t(w))))))
r2: h(x,c(y,z),t(w)) -> h(c(s(y),x),z,t(c(t(w),w)))
r3: h(c(s(x),c(s(|0|()),y)),z,t(x)) -> h(y,c(s(|0|()),c(x,z)),t(t(c(x,s(x)))))
r4: t(t(x)) -> t(c(t(x),x))
r5: t(x) -> x
r6: t(x) -> c(|0|(),c(|0|(),c(|0|(),c(|0|(),c(|0|(),x)))))

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6}


-- Reduction pair.

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

p1: h#(c(x,y),c(s(z),z),t(w)) -> h#(z,c(y,x),t(t(c(x,c(y,t(w))))))
p2: h#(c(s(x),c(s(|0|()),y)),z,t(x)) -> h#(y,c(s(|0|()),c(x,z)),t(t(c(x,s(x)))))
p3: h#(x,c(y,z),t(w)) -> h#(c(s(y),x),z,t(c(t(w),w)))

and R consists of:

r1: h(c(x,y),c(s(z),z),t(w)) -> h(z,c(y,x),t(t(c(x,c(y,t(w))))))
r2: h(x,c(y,z),t(w)) -> h(c(s(y),x),z,t(c(t(w),w)))
r3: h(c(s(x),c(s(|0|()),y)),z,t(x)) -> h(y,c(s(|0|()),c(x,z)),t(t(c(x,s(x)))))
r4: t(t(x)) -> t(c(t(x),x))
r5: t(x) -> x
r6: t(x) -> c(|0|(),c(|0|(),c(|0|(),c(|0|(),c(|0|(),x)))))

The set of usable rules consists of

  r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        h#_A(x1,x2,x3) = x1 + x2
        c_A(x1,x2) = x1 + x2 + (1,1)
        s_A(x1) = x1 + (0,1)
        t_A(x1) = ((1,1),(1,1)) x1 + (11,3)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        h#_A(x1,x2,x3) = ((1,0),(1,0)) x1 + ((0,1),(0,1)) x2
        c_A(x1,x2) = x1 + x2 + (1,3)
        s_A(x1) = ((0,1),(0,0)) x1 + (1,1)
        t_A(x1) = x1 + (1,1)
        |0|_A() = (1,4)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        h#_A(x1,x2,x3) = ((1,1),(0,0)) x1
        c_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
        s_A(x1) = (1,1)
        t_A(x1) = (1,1)
        |0|_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3

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