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. lexicographic path order with precedence:
    
      precedence:
      
        bot > new > old > top > ok > check > top#
      
      argument filter:
    
        pi(top#) = [1]
        pi(ok) = 1
        pi(new) = 1
        pi(check) = 1
        pi(old) = 1
        pi(top) = 1
        pi(bot) = []
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        top#_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1
        ok_A(x1) = x1 + (1,1,0,1)
        new_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,0,0)) x1 + (0,0,1,1)
        check_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,0)) x1 + (1,1,1,2)
        old_A(x1) = x1 + (1,1,0,3)
        top_A(x1) = x1
        bot_A() = (0,0,2,2)
    

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. lexicographic path order with precedence:
    
      precedence:
      
        old > bot > check > top > ok > top# > new
      
      argument filter:
    
        pi(top#) = 1
        pi(ok) = 1
        pi(new) = 1
        pi(check) = 1
        pi(top) = []
        pi(old) = [1]
        pi(bot) = []
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        top#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1
        ok_A(x1) = x1 + (0,5,1,3)
        new_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (0,0,3,4)
        check_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,0,0),(0,0,1,0)) x1 + (0,1,4,2)
        top_A(x1) = (0,0,0,0)
        old_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x1 + (5,1,0,1)
        bot_A() = (0,1,4,5)
    

The next rules are strictly ordered:

  p1

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