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)))
  plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))

  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)
p8: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z)

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: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))
r10: rand(x) -> x
r11: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p8}
  {p1}
  {p2}


-- Reduction pair.

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

p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z)
p2: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p3: plus#(s(x),y) -> plus#(x,s(y))
p4: 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: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))
r10: rand(x) -> x
r11: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = x1 + ((0,0),(1,1)) x2
        s_A(x1) = x1
        plus_A(x1,x2) = x1 + x2 + (1,1)
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
        double_A(x1) = x1
        |0|_A() = (0,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,1)) x1
        s_A(x1) = x1 + (3,3)
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        minus_A(x1,x2) = x1 + (1,1)
        double_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(1,0)) x1
        s_A(x1) = ((0,1),(0,0)) x1 + (1,2)
        plus_A(x1,x2) = (1,1)
        minus_A(x1,x2) = ((0,1),(0,0)) x1 + (1,1)
        double_A(x1) = ((0,0),(1,1)) x1 + (4,1)
        |0|_A() = (1,1)
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))
r10: rand(x) -> x
r11: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1
        minus_A(x1,x2) = x1 + (0,1)
        |0|_A() = (0,0)
        double_A(x1) = x1
        plus_A(x1,x2) = x1 + x2 + (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,1)) x1 + x2
        s_A(x1) = x1 + (2,2)
        minus_A(x1,x2) = x1 + (1,0)
        |0|_A() = (1,1)
        double_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        plus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + (3,3)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
        double_A(x1) = ((1,1),(1,1)) x1 + (0,1)
        plus_A(x1,x2) = ((1,1),(0,1)) x1 + (0,1)
        rand_A(x1) = (0,0)
    

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: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))
r10: rand(x) -> x
r11: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        double#_A(x1) = x1
        s_A(x1) = x1
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
        |0|_A() = (1,0)
        double_A(x1) = x1
        plus_A(x1,x2) = x1 + x2 + (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        double#_A(x1) = ((1,0),(1,0)) x1
        s_A(x1) = x1 + (1,1)
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
        |0|_A() = (1,1)
        double_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        plus_A(x1,x2) = ((1,0),(1,0)) x1 + (1,1)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        double#_A(x1) = (0,0)
        s_A(x1) = (1,1)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
        double_A(x1) = x1 + (2,1)
        plus_A(x1,x2) = (2,1)
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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