YES

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

  R:
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  minus(|0|(),y) -> |0|()
  minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
  if_minus(true(),s(x),y) -> |0|()
  if_minus(false(),s(x),y) -> s(minus(x,y))
  mod(|0|(),y) -> |0|()
  mod(s(x),|0|()) -> |0|()
  mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
  if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
  if_mod(false(),s(x),s(y)) -> s(x)

  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: le#(s(x),s(y)) -> le#(x,y)
p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)
p3: minus#(s(x),y) -> le#(s(x),y)
p4: if_minus#(false(),s(x),y) -> minus#(x,y)
p5: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p6: mod#(s(x),s(y)) -> le#(y,x)
p7: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p8: if_mod#(true(),s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p5, p7}
  {p2, p4}
  {p1}


-- Reduction pair.

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

p1: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p2: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      if_mod#_A(x1,x2,x3) = ((1,0,0),(0,1,0),(0,1,0)) x2 + x3
      true_A() = (1,1,1)
      s_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,7,6)
      mod#_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,0)) x1 + ((1,0,0),(0,1,0),(0,1,0)) x2 + (0,1,4)
      minus_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,1,4)
      le_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (2,1,0)
      |0|_A() = (0,1,9)
      false_A() = (1,2,0)
      if_minus_A(x1,x2,x3) = ((1,0,0),(0,1,0),(0,1,0)) x2 + (0,1,3)
      mod_A(x1,x2) = x1 + x2 + (1,1,2)
      if_mod_A(x1,x2,x3) = ((1,0,0),(0,1,0),(0,1,0)) x2 + x3 + (1,0,0)
      rand_A(x1) = x1 + (1,1,0)

The next rules are strictly ordered:

  p1, p2

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: if_minus#(false(),s(x),y) -> minus#(x,y)
p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      if_minus#_A(x1,x2,x3) = ((1,0,0),(1,1,0),(0,1,0)) x2
      false_A() = (1,2,5)
      s_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,2,1)
      minus#_A(x1,x2) = ((1,0,0),(1,1,0),(0,1,0)) x1 + (0,1,1)
      le_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x2 + (2,1,1)
      |0|_A() = (0,3,1)
      true_A() = (1,2,2)
      minus_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,1)) x1 + (0,2,3)
      if_minus_A(x1,x2,x3) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2 + (0,2,0)
      mod_A(x1,x2) = x1 + (1,0,0)
      if_mod_A(x1,x2,x3) = x2 + (1,0,0)
      rand_A(x1) = x1 + (1,1,0)

The next rules are strictly ordered:

  p1, p2

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

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,0)) x1 + ((1,0,0),(0,1,0),(0,1,0)) x2
      s_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,3,1)
      le_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x2 + (2,2,0)
      |0|_A() = (0,2,3)
      true_A() = (0,3,0)
      false_A() = (1,1,1)
      minus_A(x1,x2) = x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (0,1,4)
      if_minus_A(x1,x2,x3) = ((0,0,0),(0,0,0),(1,0,0)) x1 + x2 + (0,1,1)
      mod_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + (1,3,2)
      if_mod_A(x1,x2,x3) = ((1,0,0),(0,0,0),(0,1,0)) x2 + (1,3,1)
      rand_A(x1) = x1 + (1,1,0)

The next rules are strictly ordered:

  p1

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