YES

We show the termination of the TRS R:

  top(sent(x)) -> top(check(rest(x)))
  rest(nil()) -> sent(nil())
  rest(cons(x,y)) -> sent(y)
  check(sent(x)) -> sent(check(x))
  check(rest(x)) -> rest(check(x))
  check(cons(x,y)) -> cons(check(x),y)
  check(cons(x,y)) -> cons(x,check(y))
  check(cons(x,y)) -> cons(x,y)

-- SCC decomposition.

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

p1: top#(sent(x)) -> top#(check(rest(x)))
p2: top#(sent(x)) -> check#(rest(x))
p3: top#(sent(x)) -> rest#(x)
p4: check#(sent(x)) -> check#(x)
p5: check#(rest(x)) -> rest#(check(x))
p6: check#(rest(x)) -> check#(x)
p7: check#(cons(x,y)) -> check#(x)
p8: check#(cons(x,y)) -> check#(y)

and R consists of:

r1: top(sent(x)) -> top(check(rest(x)))
r2: rest(nil()) -> sent(nil())
r3: rest(cons(x,y)) -> sent(y)
r4: check(sent(x)) -> sent(check(x))
r5: check(rest(x)) -> rest(check(x))
r6: check(cons(x,y)) -> cons(check(x),y)
r7: check(cons(x,y)) -> cons(x,check(y))
r8: check(cons(x,y)) -> cons(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p4, p6, p7, p8}


-- Reduction pair.

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

p1: top#(sent(x)) -> top#(check(rest(x)))

and R consists of:

r1: top(sent(x)) -> top(check(rest(x)))
r2: rest(nil()) -> sent(nil())
r3: rest(cons(x,y)) -> sent(y)
r4: check(sent(x)) -> sent(check(x))
r5: check(rest(x)) -> rest(check(x))
r6: check(cons(x,y)) -> cons(check(x),y)
r7: check(cons(x,y)) -> cons(x,check(y))
r8: check(cons(x,y)) -> cons(x,y)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(0,0)) x1
        sent_A(x1) = ((1,0),(1,0)) x1 + (1,2)
        check_A(x1) = ((1,0),(1,0)) x1 + (0,1)
        rest_A(x1) = ((1,0),(1,0)) x1 + (1,2)
        nil_A() = (1,1)
        cons_A(x1,x2) = x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(0,0)) x1
        sent_A(x1) = x1 + (5,0)
        check_A(x1) = x1 + (2,1)
        rest_A(x1) = ((0,1),(0,1)) x1 + (1,0)
        nil_A() = (0,5)
        cons_A(x1,x2) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(0,0)) x1
        sent_A(x1) = (1,4)
        check_A(x1) = x1 + (1,1)
        rest_A(x1) = (2,4)
        nil_A() = (1,1)
        cons_A(x1,x2) = (0,1)
    

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#(sent(x)) -> check#(x)
p2: check#(cons(x,y)) -> check#(y)
p3: check#(cons(x,y)) -> check#(x)
p4: check#(rest(x)) -> check#(x)

and R consists of:

r1: top(sent(x)) -> top(check(rest(x)))
r2: rest(nil()) -> sent(nil())
r3: rest(cons(x,y)) -> sent(y)
r4: check(sent(x)) -> sent(check(x))
r5: check(rest(x)) -> rest(check(x))
r6: check(cons(x,y)) -> cons(check(x),y)
r7: check(cons(x,y)) -> cons(x,check(y))
r8: check(cons(x,y)) -> cons(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        check#_A(x1) = ((1,1),(0,1)) x1
        sent_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        cons_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        rest_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        check#_A(x1) = ((1,1),(1,1)) x1
        sent_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        cons_A(x1,x2) = ((1,1),(1,1)) x2 + (1,1)
        rest_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        check#_A(x1) = ((1,1),(1,1)) x1
        sent_A(x1) = ((1,0),(1,1)) x1 + (1,1)
        cons_A(x1,x2) = ((1,1),(1,1)) x2 + (1,1)
        rest_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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