YES

We show the termination of the relative TRS R/S:

  R:
  top(ok(new(x))) -> top(check(x))
  top(ok(old(x))) -> top(check(x))

  S:
  bot() -> new(bot())
  check(new(x)) -> new(check(x))
  check(old(x)) -> old(check(x))
  check(old(x)) -> ok(old(x))
  new(ok(x)) -> ok(new(x))
  old(ok(x)) -> ok(old(x))

-- SCC decomposition.

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

p1: top#(ok(new(x))) -> top#(check(x))
p2: top#(ok(old(x))) -> top#(check(x))

and R consists of:

r1: top(ok(new(x))) -> top(check(x))
r2: top(ok(old(x))) -> top(check(x))
r3: bot() -> new(bot())
r4: check(new(x)) -> new(check(x))
r5: check(old(x)) -> old(check(x))
r6: check(old(x)) -> ok(old(x))
r7: new(ok(x)) -> ok(new(x))
r8: old(ok(x)) -> ok(old(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: top#(ok(new(x))) -> top#(check(x))
p2: top#(ok(old(x))) -> top#(check(x))

and R consists of:

r1: top(ok(new(x))) -> top(check(x))
r2: top(ok(old(x))) -> top(check(x))
r3: bot() -> new(bot())
r4: check(new(x)) -> new(check(x))
r5: check(old(x)) -> old(check(x))
r6: check(old(x)) -> ok(old(x))
r7: new(ok(x)) -> ok(new(x))
r8: old(ok(x)) -> ok(old(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{0, x1 - 5}
          ok_A(x1) = max{2, x1}
          new_A(x1) = max{9, x1}
          check_A(x1) = max{3, x1}
          old_A(x1) = max{7, x1 + 4}
          top_A(x1) = max{1, x1}
          bot_A = 10
    
      precedence:
      
        check > bot > top# = ok = new = old = top
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(new) = []
        pi(check) = []
        pi(old) = [1]
        pi(top) = []
        pi(bot) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = 1
          ok_A(x1) = 11
          new_A(x1) = 11
          check_A(x1) = 11
          old_A(x1) = x1 + 21
          top_A(x1) = 22
          bot_A = 12
    
      precedence:
      
        top > check = old = bot > new > top# = ok
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(new) = []
        pi(check) = []
        pi(old) = [1]
        pi(top) = []
        pi(bot) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(ok(new(x))) -> top#(check(x))

and R consists of:

r1: top(ok(new(x))) -> top(check(x))
r2: top(ok(old(x))) -> top(check(x))
r3: bot() -> new(bot())
r4: check(new(x)) -> new(check(x))
r5: check(old(x)) -> old(check(x))
r6: check(old(x)) -> ok(old(x))
r7: new(ok(x)) -> ok(new(x))
r8: old(ok(x)) -> ok(old(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(ok(new(x))) -> top#(check(x))

and R consists of:

r1: top(ok(new(x))) -> top(check(x))
r2: top(ok(old(x))) -> top(check(x))
r3: bot() -> new(bot())
r4: check(new(x)) -> new(check(x))
r5: check(old(x)) -> old(check(x))
r6: check(old(x)) -> ok(old(x))
r7: new(ok(x)) -> ok(new(x))
r8: old(ok(x)) -> ok(old(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = x1 + (0,1)
            ok_A(x1) = x1 + (0,3)
            new_A(x1) = ((1,0),(1,1)) x1
            check_A(x1) = ((1,0),(1,1)) x1 + (0,1)
            top_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            old_A(x1) = x1 + (2,4)
            bot_A() = (0,1)
    
      precedence:
      
        check = top > old = bot > new > ok > top#
      
      partial status:
    
        pi(top#) = [1]
        pi(ok) = [1]
        pi(new) = []
        pi(check) = [1]
        pi(top) = [1]
        pi(old) = []
        pi(bot) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = x1
            ok_A(x1) = x1 + (3,1)
            new_A(x1) = (8,9)
            check_A(x1) = ((0,0),(1,0)) x1 + (8,1)
            top_A(x1) = ((1,0),(1,1)) x1 + (2,0)
            old_A(x1) = (4,2)
            bot_A() = (9,10)
    
      precedence:
      
        top# > top = old > bot > new = check > ok
      
      partial status:
    
        pi(top#) = [1]
        pi(ok) = [1]
        pi(new) = []
        pi(check) = []
        pi(top) = []
        pi(old) = []
        pi(bot) = []
    

The next rules are strictly ordered:

  p1

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