YES

Problem:
 strict:
  s(a(x)) -> s(b(x))
  b(b(x)) -> a(x)
 weak:
  f(s(x),y) -> f(x,s(y))
  s(a(x)) -> a(s(x))
  s(b(x)) -> b(s(x))
  a(s(x)) -> s(a(x))
  b(s(x)) -> s(b(x))

Proof:
 LPO Processor:
  precedence:
   f > b ~ s ~ a
  problem:
   strict:
    s(a(x)) -> s(b(x))
   weak:
    s(a(x)) -> a(s(x))
    s(b(x)) -> b(s(x))
    a(s(x)) -> s(a(x))
    b(s(x)) -> s(b(x))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [s](x0) = x0,
    
    [a](x0) = 9x0,
    
    [b](x0) = x0
   orientation:
    s(a(x)) = 9x >= x = s(b(x))
    
    s(a(x)) = 9x >= 9x = a(s(x))
    
    s(b(x)) = x >= x = b(s(x))
    
    a(s(x)) = 9x >= 9x = s(a(x))
    
    b(s(x)) = x >= x = s(b(x))
   problem:
    strict:
     
    weak:
     s(a(x)) -> a(s(x))
     s(b(x)) -> b(s(x))
     a(s(x)) -> s(a(x))
     b(s(x)) -> s(b(x))
   Qed