YES

We show the termination of the TRS R:

  rev1(|0|(),nil()) -> |0|()
  rev1(s(X),nil()) -> s(X)
  rev1(X,cons(Y,L)) -> rev1(Y,L)
  rev(nil()) -> nil()
  rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
  rev2(X,nil()) -> nil()
  rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

-- SCC decomposition.

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

p1: rev1#(X,cons(Y,L)) -> rev1#(Y,L)
p2: rev#(cons(X,L)) -> rev1#(X,L)
p3: rev#(cons(X,L)) -> rev2#(X,L)
p4: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))
p5: rev2#(X,cons(Y,L)) -> rev#(rev2(Y,L))
p6: rev2#(X,cons(Y,L)) -> rev2#(Y,L)

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p6}
  {p1}


-- Reduction pair.

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

p1: rev2#(X,cons(Y,L)) -> rev#(rev2(Y,L))
p2: rev#(cons(X,L)) -> rev2#(X,L)
p3: rev2#(X,cons(Y,L)) -> rev2#(Y,L)
p4: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev2#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (3,0)
            rev#_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            rev2_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2
            rev_A(x1) = ((1,0),(0,0)) x1
            rev1_A(x1,x2) = (0,2)
            |0|_A() = (0,1)
            nil_A() = (0,0)
            s_A(x1) = (0,1)
    
      precedence:
      
        rev2 = rev = rev1 > rev2# = cons > rev# = |0| = nil = s
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev#) = [1]
        pi(rev2) = [2]
        pi(rev) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev2#_A(x1,x2) = (0,0)
            cons_A(x1,x2) = (1,0)
            rev#_A(x1) = (0,0)
            rev2_A(x1,x2) = ((0,0),(1,0)) x2 + (3,3)
            rev_A(x1) = (2,2)
            rev1_A(x1,x2) = (0,0)
            |0|_A() = (2,2)
            nil_A() = (1,1)
            s_A(x1) = (0,0)
    
      precedence:
      
        rev2 > rev > rev1 = |0| = nil > rev2# = cons = rev# = s
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev#) = []
        pi(rev2) = []
        pi(rev) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: rev#(cons(X,L)) -> rev2#(X,L)
p2: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: rev#(cons(X,L)) -> rev2#(X,L)
p2: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev#_A(x1) = ((1,0),(1,0)) x1 + (1,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (4,2)
            rev2#_A(x1,x2) = x2 + (2,4)
            rev_A(x1) = x1
            rev2_A(x1,x2) = ((0,0),(1,0)) x1 + x2
            rev1_A(x1,x2) = ((0,0),(1,0)) x2 + (3,3)
            |0|_A() = (2,2)
            nil_A() = (1,1)
            s_A(x1) = (3,3)
    
      precedence:
      
        rev# = cons = rev2# = rev = rev2 = rev1 = |0| = nil = s
      
      partial status:
    
        pi(rev#) = []
        pi(cons) = []
        pi(rev2#) = []
        pi(rev) = []
        pi(rev2) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev#_A(x1) = (0,0)
            cons_A(x1,x2) = (3,2)
            rev2#_A(x1,x2) = ((1,0),(1,1)) x2 + (2,1)
            rev_A(x1) = (4,3)
            rev2_A(x1,x2) = ((0,0),(1,0)) x2 + (6,7)
            rev1_A(x1,x2) = (0,1)
            |0|_A() = (0,0)
            nil_A() = (1,9)
            s_A(x1) = (0,0)
    
      precedence:
      
        rev2 > rev > s > nil > rev# = cons = rev2# = rev1 = |0|
      
      partial status:
    
        pi(rev#) = []
        pi(cons) = []
        pi(rev2#) = []
        pi(rev) = []
        pi(rev2) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    

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: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: rev1#(X,cons(Y,L)) -> rev1#(Y,L)

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

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:
            rev1#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
            cons_A(x1,x2) = x1 + x2 + (2,2)
    
      precedence:
      
        rev1# = cons
      
      partial status:
    
        pi(rev1#) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev1#_A(x1,x2) = (0,0)
            cons_A(x1,x2) = (1,1)
    
      precedence:
      
        rev1# = cons
      
      partial status:
    
        pi(rev1#) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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