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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_mod#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + x3 + (4,1)
            true_A() = (1,1)
            s_A(x1) = x1 + (0,3)
            mod#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,1)
            minus_A(x1,x2) = x1
            le_A(x1,x2) = (3,2)
            |0|_A() = (0,2)
            false_A() = (1,1)
            mod_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,2)
            if_mod_A(x1,x2,x3) = x2 + ((1,0),(0,0)) x3 + (2,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        if_mod# = true = s = mod# = minus = le = |0| = false = mod = if_mod = rand
      
      partial status:
    
        pi(if_mod#) = []
        pi(true) = []
        pi(s) = []
        pi(mod#) = []
        pi(minus) = []
        pi(le) = []
        pi(|0|) = []
        pi(false) = []
        pi(mod) = []
        pi(if_mod) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_mod#_A(x1,x2,x3) = (1,0)
            true_A() = (3,0)
            s_A(x1) = (0,9)
            mod#_A(x1,x2) = ((0,0),(1,0)) x1 + (2,1)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (7,11)
            le_A(x1,x2) = (4,0)
            |0|_A() = (0,0)
            false_A() = (5,10)
            mod_A(x1,x2) = (5,1)
            if_mod_A(x1,x2,x3) = (6,10)
            rand_A(x1) = (1,1)
    
      precedence:
      
        minus > le = if_mod > false > true = mod# = rand > if_mod# = s = |0| = mod
      
      partial status:
    
        pi(if_mod#) = []
        pi(true) = []
        pi(s) = []
        pi(mod#) = []
        pi(minus) = []
        pi(le) = []
        pi(|0|) = []
        pi(false) = []
        pi(mod) = []
        pi(if_mod) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = max{3, x1 + 1, x2 + 1}
          s_A(x1) = max{2, x1}
          le_A(x1,x2) = x2 + 3
          |0|_A = 5
          true_A = 1
          false_A = 0
          minus_A(x1,x2) = max{x1, x2 + 2}
          mod_A(x1,x2) = max{10, x1 + 6, x2 + 9}
          if_mod_A(x1,x2,x3) = max{x1 + 3, x2 + 6, x3 + 9}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        le = |0| = true = false = mod = if_mod > s > le# = minus = rand
      
      partial status:
    
        pi(le#) = [1, 2]
        pi(s) = [1]
        pi(le) = [2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(minus) = [1, 2]
        pi(mod) = []
        pi(if_mod) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = max{4, x1 + 1}
          s_A(x1) = max{3, x1}
          le_A(x1,x2) = 6
          |0|_A = 0
          true_A = 3
          false_A = 3
          minus_A(x1,x2) = max{6, x1, x2}
          mod_A(x1,x2) = 5
          if_mod_A(x1,x2,x3) = 5
          rand_A(x1) = 1
    
      precedence:
      
        le# = s = le = |0| = true = false = minus = mod = if_mod = rand
      
      partial status:
    
        pi(le#) = [1]
        pi(s) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(minus) = [1, 2]
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{3, x1 + 1, x2}
          s_A(x1) = max{2, x1}
          le_A(x1,x2) = x2 + 3
          |0|_A = 5
          true_A = 1
          false_A = 0
          minus_A(x1,x2) = max{x1, x2 + 2}
          mod_A(x1,x2) = max{10, x1 + 6, x2 + 9}
          if_mod_A(x1,x2,x3) = max{x1 + 3, x2 + 6, x3 + 9}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        le = |0| = true = false = mod = if_mod > s > minus# = minus = rand
      
      partial status:
    
        pi(minus#) = [1, 2]
        pi(s) = [1]
        pi(le) = [2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(minus) = [1, 2]
        pi(mod) = []
        pi(if_mod) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{4, x1 + 1}
          s_A(x1) = max{3, x1}
          le_A(x1,x2) = 7
          |0|_A = 0
          true_A = 5
          false_A = 5
          minus_A(x1,x2) = max{6, x1, x2}
          mod_A(x1,x2) = 5
          if_mod_A(x1,x2,x3) = 5
          rand_A(x1) = 1
    
      precedence:
      
        minus# = s = le = |0| = true = false = minus = mod = if_mod = rand
      
      partial status:
    
        pi(minus#) = [1]
        pi(s) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(minus) = [1, 2]
        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.