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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          rev2#_A(x1,x2) = ((1,0),(1,0)) x2 + (2,0)
          cons_A(x1,x2) = ((1,0),(1,0)) x2 + (5,0)
          rev#_A(x1) = x1 + (1,1)
          rev2_A(x1,x2) = ((1,0),(1,0)) x2 + (0,2)
          rev_A(x1) = ((1,0),(1,0)) x1 + (0,1)
          rev1_A(x1,x2) = (4,7)
          |0|_A() = (1,2)
          nil_A() = (2,1)
          s_A(x1) = (3,8)
  
    precedence:
    
      rev2 > rev > cons > rev2# = rev# > rev1 = |0| = nil = 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:

  p3

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


-- 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)) -> 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          rev2#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,4)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (3,1)
          rev#_A(x1) = ((1,0),(0,0)) x1 + (2,8)
          rev2_A(x1,x2) = ((1,0),(0,0)) x2 + (0,7)
          rev_A(x1) = ((1,0),(0,0)) x1 + (0,6)
          rev1_A(x1,x2) = (4,0)
          |0|_A() = (1,6)
          nil_A() = (0,5)
          s_A(x1) = (1,1)
  
    precedence:
    
      rev2# = cons = rev# = rev2 = rev = rev1 = |0| = nil = 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

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          rev#_A(x1) = ((1,0),(1,0)) x1 + (2,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (3,3)
          rev2#_A(x1,x2) = ((1,0),(1,0)) x2 + (4,2)
          rev_A(x1) = ((1,0),(0,0)) x1 + (0,7)
          rev2_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,8)
          rev1_A(x1,x2) = ((1,0),(0,0)) x2 + (2,4)
          |0|_A() = (1,5)
          nil_A() = (2,6)
          s_A(x1) = (3,5)
  
    precedence:
    
      cons = rev = rev2 > rev2# = rev1 = s > |0| = nil > rev#
    
    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:

  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) = ((1,0),(1,1)) x1 + x2 + (2,2)
  
    precedence:
    
      cons > rev1#
    
    partial status:
  
      pi(rev1#) = [2]
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

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