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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          if_mod#_A(x1,x2,x3) = x2
          true_A() = 0
          s_A(x1) = x1
          mod#_A(x1,x2) = x1
          minus_A(x1,x2) = x1
          le_A(x1,x2) = x1 + 3
          |0|_A() = 1
          false_A() = 2
          mod_A(x1,x2) = x1 + x2 + 1
          if_mod_A(x1,x2,x3) = x2 + x3 + 1
          rand_A(x1) = x1 + 1
  
    precedence:
    
      |0| = mod = if_mod = rand > s > minus > mod# > if_mod# = true > le = false
    
    partial status:
  
      pi(if_mod#) = [2]
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = [1]
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(false) = []
      pi(mod) = [1, 2]
      pi(if_mod) = [2, 3]
      pi(rand) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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))

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:

  (no SCCs)

-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          le#_A(x1,x2) = x2
          s_A(x1) = x1
          le_A(x1,x2) = x1 + x2 + 1
          |0|_A() = 2
          true_A() = 1
          false_A() = 1
          minus_A(x1,x2) = x1
          mod_A(x1,x2) = x1 + x2 + 2
          if_mod_A(x1,x2,x3) = x2 + x3 + 2
          rand_A(x1) = x1 + 1
  
    precedence:
    
      le = |0| = false > true = minus = mod = if_mod > le# = s > rand
    
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = [1]
      pi(mod) = []
      pi(if_mod) = []
      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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          minus#_A(x1,x2) = x2
          s_A(x1) = x1
          le_A(x1,x2) = x1 + 4
          |0|_A() = 2
          true_A() = 1
          false_A() = 3
          minus_A(x1,x2) = x1
          mod_A(x1,x2) = x1 + x2 + 5
          if_mod_A(x1,x2,x3) = x2 + x3 + 5
          rand_A(x1) = x1 + 1
  
    precedence:
    
      minus# = s = le = |0| = true = false = minus = mod = if_mod > rand
    
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = [1]
      pi(mod) = [1, 2]
      pi(if_mod) = [2, 3]
      pi(rand) = []

The next rules are strictly ordered:

  p1

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