YES

We show the termination of the TRS R:

  rev(nil()) -> nil()
  rev(.(x,y)) -> ++(rev(y),.(x,nil()))
  car(.(x,y)) -> x
  cdr(.(x,y)) -> y
  null(nil()) -> true()
  null(.(x,y)) -> false()
  ++(nil(),y) -> y
  ++(.(x,y),z) -> .(x,++(y,z))

-- SCC decomposition.

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

p1: rev#(.(x,y)) -> ++#(rev(y),.(x,nil()))
p2: rev#(.(x,y)) -> rev#(y)
p3: ++#(.(x,y),z) -> ++#(y,z)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(.(x,y)) -> ++(rev(y),.(x,nil()))
r3: car(.(x,y)) -> x
r4: cdr(.(x,y)) -> y
r5: null(nil()) -> true()
r6: null(.(x,y)) -> false()
r7: ++(nil(),y) -> y
r8: ++(.(x,y),z) -> .(x,++(y,z))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

p1: rev#(.(x,y)) -> rev#(y)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(.(x,y)) -> ++(rev(y),.(x,nil()))
r3: car(.(x,y)) -> x
r4: cdr(.(x,y)) -> y
r5: null(nil()) -> true()
r6: null(.(x,y)) -> false()
r7: ++(nil(),y) -> y
r8: ++(.(x,y),z) -> .(x,++(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      . > rev#
    
    argument filter:
  
      pi(rev#) = [1]
      pi(.) = [2]

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: ++#(.(x,y),z) -> ++#(y,z)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(.(x,y)) -> ++(rev(y),.(x,nil()))
r3: car(.(x,y)) -> x
r4: cdr(.(x,y)) -> y
r5: null(nil()) -> true()
r6: null(.(x,y)) -> false()
r7: ++(nil(),y) -> y
r8: ++(.(x,y),z) -> .(x,++(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      . > ++#
    
    argument filter:
  
      pi(++#) = [1]
      pi(.) = [1, 2]

The next rules are strictly ordered:

  p1

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