YES

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

  R:
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  double(|0|()) -> |0|()
  double(s(x)) -> s(s(double(x)))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  plus(s(x),y) -> plus(x,s(y))
  plus(s(x),y) -> s(plus(minus(x,y),double(y)))

  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: minus#(s(x),s(y)) -> minus#(x,y)
p2: double#(s(x)) -> double#(x)
p3: plus#(s(x),y) -> plus#(x,y)
p4: plus#(s(x),y) -> plus#(x,s(y))
p5: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p6: plus#(s(x),y) -> minus#(x,y)
p7: plus#(s(x),y) -> double#(y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5}
  {p1}
  {p2}


-- Reduction pair.

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

p1: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p2: plus#(s(x),y) -> plus#(x,s(y))
p3: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = x1
            s_A(x1) = x1
            minus_A(x1,x2) = x1
            double_A(x1) = x1
            |0|_A() = (1,1)
            plus_A(x1,x2) = x1 + x2 + (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        rand > double > plus > s > minus = |0| > plus#
      
      partial status:
    
        pi(plus#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(double) = [1]
        pi(|0|) = []
        pi(plus) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,3)
            s_A(x1) = x1 + (3,2)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (0,1)
            double_A(x1) = (2,0)
            |0|_A() = (1,1)
            plus_A(x1,x2) = ((1,0),(1,0)) x1 + (0,3)
            rand_A(x1) = (1,1)
    
      precedence:
      
        minus = rand > double = |0| = plus > s > plus#
      
      partial status:
    
        pi(plus#) = [1]
        pi(s) = []
        pi(minus) = []
        pi(double) = []
        pi(|0|) = []
        pi(plus) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(x),y) -> plus#(x,s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus#(s(x),y) -> plus#(x,s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = x1
            s_A(x1) = x1
            minus_A(x1,x2) = x1
            |0|_A() = (1,1)
            double_A(x1) = x1 + (0,2)
            plus_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        |0| = double = plus > s > minus > plus# = rand
      
      partial status:
    
        pi(plus#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(double) = [1]
        pi(plus) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = ((1,0),(1,1)) x1
            s_A(x1) = (3,3)
            minus_A(x1,x2) = x1 + (1,5)
            |0|_A() = (0,1)
            double_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            plus_A(x1,x2) = ((1,0),(0,0)) x1 + (1,4)
            rand_A(x1) = (0,0)
    
      precedence:
      
        |0| = double = plus > plus# = s > minus = rand
      
      partial status:
    
        pi(plus#) = [1]
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(double) = []
        pi(plus) = []
        pi(rand) = []
    

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: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = x1
            s_A(x1) = x1
            minus_A(x1,x2) = x1
            |0|_A() = (1,1)
            double_A(x1) = x1 + (0,2)
            plus_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        double > |0| = plus > minus# = s > minus > rand
      
      partial status:
    
        pi(minus#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(double) = [1]
        pi(plus) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = x1 + (0,1)
            s_A(x1) = x1 + (0,2)
            minus_A(x1,x2) = (0,0)
            |0|_A() = (1,1)
            double_A(x1) = (1,0)
            plus_A(x1,x2) = x1 + (2,3)
            rand_A(x1) = (1,3)
    
      precedence:
      
        |0| = double = plus > minus# = s = minus = rand
      
      partial status:
    
        pi(minus#) = [1]
        pi(s) = [1]
        pi(minus) = []
        pi(|0|) = []
        pi(double) = []
        pi(plus) = [1]
        pi(rand) = []
    

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: double#(s(x)) -> double#(x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            double#_A(x1) = x1
            s_A(x1) = x1
            minus_A(x1,x2) = x1
            |0|_A() = (1,1)
            double_A(x1) = x1 + (0,1)
            plus_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,2)
            rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
      precedence:
      
        plus > |0| = double > s = rand > double# = minus
      
      partial status:
    
        pi(double#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(double) = [1]
        pi(plus) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            double#_A(x1) = ((0,0),(1,0)) x1
            s_A(x1) = x1 + (2,1)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + (3,2)
            |0|_A() = (1,1)
            double_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            plus_A(x1,x2) = (0,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        rand > |0| = plus > double# > minus > s = double
      
      partial status:
    
        pi(double#) = []
        pi(s) = [1]
        pi(minus) = []
        pi(|0|) = []
        pi(double) = [1]
        pi(plus) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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