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(x,|0|()) -> x
  minus(s(x),s(y)) -> 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),s(y)) -> minus#(x,y)
p3: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p4: mod#(s(x),s(y)) -> le#(y,x)
p5: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p6: 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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)
r11: rand(x) -> x
r12: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p3, p5}
  {p1}
  {p2}


-- 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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)
r11: rand(x) -> x
r12: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_mod#_A(x1,x2,x3) = ((1,0),(1,0)) x2 + ((1,0),(1,0)) x3
        true_A() = (1,1)
        s_A(x1) = x1
        mod#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2
        minus_A(x1,x2) = x1
        le_A(x1,x2) = ((1,1),(1,1)) x1 + (2,1)
        |0|_A() = (1,1)
        false_A() = (1,1)
        mod_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (1,0)
        if_mod_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(1,0)) x3 + (1,0)
        rand_A(x1) = ((1,0),(1,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_mod#_A(x1,x2,x3) = ((1,1),(1,1)) x2 + x3
        true_A() = (1,5)
        s_A(x1) = ((1,1),(1,1)) x1 + (0,3)
        mod#_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,0)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (0,1)
        le_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,2)
        false_A() = (5,5)
        mod_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (2,0)
        if_mod_A(x1,x2,x3) = ((1,0),(1,0)) x2 + ((0,0),(1,0)) x3 + (2,0)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_mod#_A(x1,x2,x3) = ((0,1),(0,1)) x2 + ((1,1),(1,0)) x3
        true_A() = (1,1)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,3)
        mod#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (1,0)
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
        le_A(x1,x2) = (2,1)
        |0|_A() = (1,1)
        false_A() = (3,2)
        mod_A(x1,x2) = ((0,1),(0,0)) x1 + (1,1)
        if_mod_A(x1,x2,x3) = x2 + (2,1)
        rand_A(x1) = (0,0)
    

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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)
r11: rand(x) -> x
r12: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        le#_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(0,0)) x2
        s_A(x1) = x1
        le_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (1,1)
        |0|_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,1)
        minus_A(x1,x2) = x1
        mod_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,1)
        if_mod_A(x1,x2,x3) = x2 + ((1,1),(0,0)) x3 + (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        le#_A(x1,x2) = ((0,1),(0,0)) x1
        s_A(x1) = ((0,0),(1,1)) x1 + (7,3)
        le_A(x1,x2) = ((1,1),(0,0)) x1 + (1,1)
        |0|_A() = (3,1)
        true_A() = (1,2)
        false_A() = (1,2)
        minus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
        mod_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (1,0)
        if_mod_A(x1,x2,x3) = ((0,1),(0,1)) x2 + ((0,1),(0,1)) x3
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        le#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
        le_A(x1,x2) = x1 + (2,1)
        |0|_A() = (2,1)
        true_A() = (1,1)
        false_A() = (3,1)
        minus_A(x1,x2) = ((0,1),(1,1)) x1 + (1,0)
        mod_A(x1,x2) = (1,0)
        if_mod_A(x1,x2,x3) = (0,0)
        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: minus#(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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)
r11: rand(x) -> x
r12: rand(x) -> rand(s(x))

The set of usable rules consists of

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

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) = ((1,1),(0,0)) x1
        le_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,1)
        |0|_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,1)
        minus_A(x1,x2) = ((1,0),(1,1)) x1
        mod_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (1,1)
        if_mod_A(x1,x2,x3) = x2 + ((1,0),(1,0)) x3 + (1,1)
        rand_A(x1) = ((1,1),(0,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        s_A(x1) = ((0,1),(1,1)) x1 + (3,1)
        le_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2 + (1,1)
        |0|_A() = (1,1)
        true_A() = (2,4)
        false_A() = (1,6)
        minus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
        mod_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,0)
        if_mod_A(x1,x2,x3) = ((1,1),(0,0)) x2 + x3
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((0,0),(1,1)) x1
        s_A(x1) = (1,3)
        le_A(x1,x2) = x1 + x2 + (1,1)
        |0|_A() = (4,5)
        true_A() = (6,2)
        false_A() = (7,1)
        minus_A(x1,x2) = (1,1)
        mod_A(x1,x2) = ((0,1),(0,1)) x1
        if_mod_A(x1,x2,x3) = (2,3)
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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