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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      rev2#_A(x1,x2) = x2 + (1,0)
      cons_A(x1,x2) = ((1,1),(1,0)) x2 + (0,1)
      rev#_A(x1) = ((0,1),(0,0)) x1
      rev2_A(x1,x2) = x2
      rev_A(x1) = x1
      rev1_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1)
      |0|_A() = (1,1)
      nil_A() = (1,1)
      s_A(x1) = (1,1)

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)) -> rev2#(Y,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: rev#(cons(X,L)) -> rev2#(X,L)
p2: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))
p3: 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 set of usable rules consists of

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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      rev#_A(x1) = ((0,1),(0,0)) x1
      cons_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (1,2)
      rev2#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,0)
      rev_A(x1) = ((0,1),(0,1)) x1
      rev2_A(x1,x2) = x1 + ((0,1),(0,1)) x2
      rev1_A(x1,x2) = x2 + (1,1)
      |0|_A() = (1,1)
      nil_A() = (1,1)
      s_A(x1) = (1,1)

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: 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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      rev1#_A(x1,x2) = ((0,1),(1,1)) x2
      cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (0,1)

The next rules are strictly ordered:

  p1

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