YES

We show the termination of the relative TRS R/S:

  R:
  rev(nil()) -> nil()
  rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
  rev1(|0|(),nil()) -> |0|()
  rev1(s(x),nil()) -> s(x)
  rev1(x,cons(y,l)) -> rev1(y,l)
  rev2(x,nil()) -> nil()
  rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

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

p1: rev#(cons(x,l)) -> rev1#(x,l)
p2: rev#(cons(x,l)) -> rev2#(x,l)
p3: rev1#(x,cons(y,l)) -> rev1#(y,l)
p4: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
p5: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5}
  {p3}


-- Reduction pair.

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

p1: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
p2: rev#(cons(x,l)) -> rev2#(x,l)
p3: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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 + (2,4)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (3,2)
            rev#_A(x1) = ((1,0),(0,0)) x1 + (1,3)
            rev2_A(x1,x2) = x2 + (0,3)
            rev_A(x1) = x1 + (0,2)
            nil_A() = (0,2)
            rev1_A(x1,x2) = ((0,0),(1,0)) x2 + (1,5)
            |0|_A() = (0,1)
            s_A(x1) = (0,3)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,4)
    
      precedence:
      
        rev2 = rev > cons = rev# = nil = |0| > rev2# = s > rev1 = rand
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev#) = []
        pi(rev2) = [2]
        pi(rev) = []
        pi(nil) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rev2#_A(x1,x2) = (3,3)
            cons_A(x1,x2) = (0,1)
            rev#_A(x1) = (4,2)
            rev2_A(x1,x2) = (2,2)
            rev_A(x1) = (1,3)
            nil_A() = (0,0)
            rev1_A(x1,x2) = (0,0)
            |0|_A() = (0,0)
            s_A(x1) = (1,1)
            rand_A(x1) = (0,0)
    
      precedence:
      
        rev# > rev2# = cons = rev2 = |0| = s = rand > rev = nil > rev1
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev#) = []
        pi(rev2) = []
        pi(rev) = []
        pi(nil) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimal 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)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: rev1#(x,cons(y,l)) -> rev1#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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) = ((0,0),(1,0)) x2 + (1,1)
            cons_A(x1,x2) = ((1,0),(1,0)) x2 + (2,2)
            rev_A(x1) = x1 + (0,1)
            nil_A() = (1,2)
            rev1_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,0)
            rev2_A(x1,x2) = ((1,0),(1,1)) x2
            |0|_A() = (0,1)
            s_A(x1) = (0,0)
            rand_A(x1) = x1
    
      precedence:
      
        rev1# > rand > rev2 > cons = rev = |0| > nil > rev1 = s
      
      partial status:
    
        pi(rev1#) = []
        pi(cons) = []
        pi(rev) = [1]
        pi(nil) = []
        pi(rev1) = [2]
        pi(rev2) = [2]
        pi(|0|) = []
        pi(s) = []
        pi(rand) = [1]
    
    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) = (4,2)
            rev_A(x1) = (5,3)
            nil_A() = (1,0)
            rev1_A(x1,x2) = (0,1)
            rev2_A(x1,x2) = ((0,0),(1,0)) x2 + (6,0)
            |0|_A() = (0,0)
            s_A(x1) = (0,0)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        rand > |0| > cons = rev2 > rev1# = rev = nil = rev1 = s
      
      partial status:
    
        pi(rev1#) = []
        pi(cons) = []
        pi(rev) = []
        pi(nil) = []
        pi(rev1) = []
        pi(rev2) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = [1]
    

The next rules are strictly ordered:

  p1

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