YES

We show the termination of the TRS R:

  rec(rec(x)) -> sent(rec(x))
  rec(sent(x)) -> sent(rec(x))
  rec(no(x)) -> sent(rec(x))
  rec(bot()) -> up(sent(bot()))
  rec(up(x)) -> up(rec(x))
  sent(up(x)) -> up(sent(x))
  no(up(x)) -> up(no(x))
  top(rec(up(x))) -> top(check(rec(x)))
  top(sent(up(x))) -> top(check(rec(x)))
  top(no(up(x))) -> top(check(rec(x)))
  check(up(x)) -> up(check(x))
  check(sent(x)) -> sent(check(x))
  check(rec(x)) -> rec(check(x))
  check(no(x)) -> no(check(x))
  check(no(x)) -> no(x)

-- SCC decomposition.

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

p1: rec#(rec(x)) -> sent#(rec(x))
p2: rec#(sent(x)) -> sent#(rec(x))
p3: rec#(sent(x)) -> rec#(x)
p4: rec#(no(x)) -> sent#(rec(x))
p5: rec#(no(x)) -> rec#(x)
p6: rec#(bot()) -> sent#(bot())
p7: rec#(up(x)) -> rec#(x)
p8: sent#(up(x)) -> sent#(x)
p9: no#(up(x)) -> no#(x)
p10: top#(rec(up(x))) -> top#(check(rec(x)))
p11: top#(rec(up(x))) -> check#(rec(x))
p12: top#(rec(up(x))) -> rec#(x)
p13: top#(sent(up(x))) -> top#(check(rec(x)))
p14: top#(sent(up(x))) -> check#(rec(x))
p15: top#(sent(up(x))) -> rec#(x)
p16: top#(no(up(x))) -> top#(check(rec(x)))
p17: top#(no(up(x))) -> check#(rec(x))
p18: top#(no(up(x))) -> rec#(x)
p19: check#(up(x)) -> check#(x)
p20: check#(sent(x)) -> sent#(check(x))
p21: check#(sent(x)) -> check#(x)
p22: check#(rec(x)) -> rec#(check(x))
p23: check#(rec(x)) -> check#(x)
p24: check#(no(x)) -> no#(check(x))
p25: check#(no(x)) -> check#(x)

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The estimated dependency graph contains the following SCCs:

  {p10, p13, p16}
  {p19, p21, p23, p25}
  {p3, p5, p7}
  {p8}
  {p9}


-- Reduction pair.

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

p1: top#(no(up(x))) -> top#(check(rec(x)))
p2: top#(sent(up(x))) -> top#(check(rec(x)))
p3: top#(rec(up(x))) -> top#(check(rec(x)))

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r11, r12, r13, r14, r15

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) = ((0,0),(1,0)) x1 + (3,9)
            no_A(x1) = ((0,0),(1,0)) x1 + (4,4)
            up_A(x1) = (2,1)
            check_A(x1) = ((1,0),(1,1)) x1 + (0,5)
            rec_A(x1) = (2,3)
            sent_A(x1) = ((1,0),(1,0)) x1
            bot_A() = (2,2)
    
      precedence:
      
        no = check = rec > top# = sent > bot > up
      
      partial status:
    
        pi(top#) = []
        pi(no) = []
        pi(up) = []
        pi(check) = [1]
        pi(rec) = []
        pi(sent) = []
        pi(bot) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = (0,0)
            no_A(x1) = (5,4)
            up_A(x1) = (1,1)
            check_A(x1) = ((0,0),(1,0)) x1 + (4,5)
            rec_A(x1) = (3,3)
            sent_A(x1) = (2,2)
            bot_A() = (0,2)
    
      precedence:
      
        bot > top# = check > sent > no = up = rec
      
      partial status:
    
        pi(top#) = []
        pi(no) = []
        pi(up) = []
        pi(check) = []
        pi(rec) = []
        pi(sent) = []
        pi(bot) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(sent(up(x))) -> top#(check(rec(x)))
p2: top#(rec(up(x))) -> top#(check(rec(x)))

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: top#(sent(up(x))) -> top#(check(rec(x)))
p2: top#(rec(up(x))) -> top#(check(rec(x)))

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r11, r12, r13, r14, r15

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
            sent_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            up_A(x1) = x1 + (0,1)
            check_A(x1) = x1
            rec_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            no_A(x1) = ((1,0),(1,1)) x1 + (3,4)
            bot_A() = (1,1)
    
      precedence:
      
        check = rec = no > top# = sent > up = bot
      
      partial status:
    
        pi(top#) = []
        pi(sent) = [1]
        pi(up) = [1]
        pi(check) = [1]
        pi(rec) = [1]
        pi(no) = [1]
        pi(bot) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = (0,0)
            sent_A(x1) = ((1,0),(1,1)) x1 + (4,2)
            up_A(x1) = ((1,0),(1,1)) x1 + (6,5)
            check_A(x1) = ((1,0),(1,1)) x1 + (3,1)
            rec_A(x1) = ((1,0),(1,1)) x1 + (4,3)
            no_A(x1) = ((1,0),(1,1)) x1 + (5,4)
            bot_A() = (1,1)
    
      precedence:
      
        bot > up = check = rec > sent > top# = no
      
      partial status:
    
        pi(top#) = []
        pi(sent) = []
        pi(up) = []
        pi(check) = []
        pi(rec) = []
        pi(no) = [1]
        pi(bot) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(sent(up(x))) -> top#(check(rec(x)))

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(sent(up(x))) -> top#(check(rec(x)))

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r11, r12, r13, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = ((1,0),(0,0)) x1
            sent_A(x1) = ((1,1),(0,0)) x1 + (1,0)
            up_A(x1) = x1 + (2,0)
            check_A(x1) = ((1,0),(0,0)) x1
            rec_A(x1) = ((1,0),(0,0)) x1 + (3,0)
            no_A(x1) = ((1,0),(0,0)) x1 + (5,0)
            bot_A() = (1,0)
    
      precedence:
      
        top# = sent = up = check = rec = no = bot
      
      partial status:
    
        pi(top#) = []
        pi(sent) = []
        pi(up) = []
        pi(check) = []
        pi(rec) = []
        pi(no) = []
        pi(bot) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = ((1,1),(0,0)) x1 + (1,1)
            sent_A(x1) = x1 + (7,0)
            up_A(x1) = x1
            check_A(x1) = ((1,0),(0,0)) x1 + (3,0)
            rec_A(x1) = ((1,1),(0,1)) x1 + (2,0)
            no_A(x1) = (1,0)
            bot_A() = (1,6)
    
      precedence:
      
        check > sent = rec = no > top# = up = bot
      
      partial status:
    
        pi(top#) = []
        pi(sent) = []
        pi(up) = []
        pi(check) = []
        pi(rec) = [1]
        pi(no) = []
        pi(bot) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: check#(no(x)) -> check#(x)
p2: check#(rec(x)) -> check#(x)
p3: check#(sent(x)) -> check#(x)
p4: check#(up(x)) -> check#(x)

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            check#_A(x1) = x1 + (2,2)
            no_A(x1) = x1 + (3,3)
            rec_A(x1) = ((1,0),(1,1)) x1 + (3,3)
            sent_A(x1) = x1 + (1,1)
            up_A(x1) = x1 + (1,1)
    
      precedence:
      
        check# = no = rec = sent = up
      
      partial status:
    
        pi(check#) = []
        pi(no) = []
        pi(rec) = []
        pi(sent) = []
        pi(up) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            check#_A(x1) = ((1,0),(0,0)) x1
            no_A(x1) = (1,1)
            rec_A(x1) = (1,1)
            sent_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            up_A(x1) = (1,1)
    
      precedence:
      
        no > up > check# = rec = sent
      
      partial status:
    
        pi(check#) = []
        pi(no) = []
        pi(rec) = []
        pi(sent) = []
        pi(up) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

p1: rec#(up(x)) -> rec#(x)
p2: rec#(no(x)) -> rec#(x)
p3: rec#(sent(x)) -> rec#(x)

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rec#_A(x1) = x1
            up_A(x1) = x1 + (1,1)
            no_A(x1) = x1 + (1,1)
            sent_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        no = sent > rec# = up
      
      partial status:
    
        pi(rec#) = [1]
        pi(up) = [1]
        pi(no) = [1]
        pi(sent) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rec#_A(x1) = ((1,0),(1,0)) x1
            up_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            no_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            sent_A(x1) = (1,1)
    
      precedence:
      
        rec# = up = no > sent
      
      partial status:
    
        pi(rec#) = []
        pi(up) = []
        pi(no) = []
        pi(sent) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: sent#(up(x)) -> sent#(x)

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sent#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            up_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        sent# = up
      
      partial status:
    
        pi(sent#) = []
        pi(up) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sent#_A(x1) = ((1,0),(0,0)) x1
            up_A(x1) = (1,1)
    
      precedence:
      
        up > sent#
      
      partial status:
    
        pi(sent#) = []
        pi(up) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: no#(up(x)) -> no#(x)

and R consists of:

r1: rec(rec(x)) -> sent(rec(x))
r2: rec(sent(x)) -> sent(rec(x))
r3: rec(no(x)) -> sent(rec(x))
r4: rec(bot()) -> up(sent(bot()))
r5: rec(up(x)) -> up(rec(x))
r6: sent(up(x)) -> up(sent(x))
r7: no(up(x)) -> up(no(x))
r8: top(rec(up(x))) -> top(check(rec(x)))
r9: top(sent(up(x))) -> top(check(rec(x)))
r10: top(no(up(x))) -> top(check(rec(x)))
r11: check(up(x)) -> up(check(x))
r12: check(sent(x)) -> sent(check(x))
r13: check(rec(x)) -> rec(check(x))
r14: check(no(x)) -> no(check(x))
r15: check(no(x)) -> no(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            no#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            up_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        no# = up
      
      partial status:
    
        pi(no#) = []
        pi(up) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            no#_A(x1) = ((1,0),(0,0)) x1
            up_A(x1) = (1,1)
    
      precedence:
      
        up > no#
      
      partial status:
    
        pi(no#) = []
        pi(up) = []
    

The next rules are strictly ordered:

  p1

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