YES

Problem:
 strict:
  n(s(x1)) -> s(x1)
  o(s(x1)) -> s(x1)
 weak:
  t(x1) -> t(c(n(x1)))
  c(n(x1)) -> n(c(x1))
  c(o(x1)) -> o(c(x1))
  c(o(x1)) -> o(x1)

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = x0,
   
   [n](x0) = x0,
   
   [t](x0) = 1x0,
   
   [s](x0) = x0,
   
   [o](x0) = 7x0
  orientation:
   n(s(x1)) = x1 >= x1 = s(x1)
   
   o(s(x1)) = 7x1 >= x1 = s(x1)
   
   t(x1) = 1x1 >= 1x1 = t(c(n(x1)))
   
   c(n(x1)) = x1 >= x1 = n(c(x1))
   
   c(o(x1)) = 7x1 >= 7x1 = o(c(x1))
   
   c(o(x1)) = 7x1 >= 7x1 = o(x1)
  problem:
   strict:
    n(s(x1)) -> s(x1)
   weak:
    t(x1) -> t(c(n(x1)))
    c(n(x1)) -> n(c(x1))
    c(o(x1)) -> o(c(x1))
    c(o(x1)) -> o(x1)
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 0]  
    [c](x0) = [0 0]x0,
    
              [1 1]  
    [n](x0) = [0 1]x0,
    
              [2 2]  
    [t](x0) = [1 1]x0,
    
              [1 1]     [0]
    [s](x0) = [2 0]x0 + [2],
    
              [2 2]  
    [o](x0) = [0 0]x0
   orientation:
               [3 1]     [2]    [1 1]     [0]        
    n(s(x1)) = [2 0]x1 + [2] >= [2 0]x1 + [2] = s(x1)
    
            [2 2]      [2 2]                
    t(x1) = [1 1]x1 >= [1 1]x1 = t(c(n(x1)))
    
               [1 1]      [1 0]             
    c(n(x1)) = [0 0]x1 >= [0 0]x1 = n(c(x1))
    
               [2 2]      [2 0]             
    c(o(x1)) = [0 0]x1 >= [0 0]x1 = o(c(x1))
    
               [2 2]      [2 2]          
    c(o(x1)) = [0 0]x1 >= [0 0]x1 = o(x1)
   problem:
    strict:
     
    weak:
     t(x1) -> t(c(n(x1)))
     c(n(x1)) -> n(c(x1))
     c(o(x1)) -> o(c(x1))
     c(o(x1)) -> o(x1)
   Qed