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)
  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(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#(x,s(y)) -> pred#(minus(x,y))
p3: minus#(x,s(y)) -> minus#(x,y)
p4: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p5: mod#(s(x),s(y)) -> le#(y,x)
p6: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p7: 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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)
r12: rand(x) -> x
r13: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p4, p6}
  {p1}
  {p3}


-- 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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)
r12: rand(x) -> x
r13: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            if_mod#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + x2 + ((1,0),(1,0)) x3 + (1,2)
            true_A() = (2,1)
            s_A(x1) = ((1,1),(0,0)) x1
            mod#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (3,2)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,3)
            le_A(x1,x2) = ((0,0),(1,1)) x1 + (2,4)
            |0|_A() = (3,1)
            false_A() = (1,1)
            pred_A(x1) = ((1,0),(1,0)) x1 + (0,1)
            mod_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,3)
            if_mod_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(1,1)) x3 + (1,3)
            rand_A(x1) = ((1,1),(0,1)) x1 + (1,1)
    
      precedence:
      
        if_mod# = true = s = mod# = minus = le = |0| = false = pred = mod = if_mod = rand
      
      partial status:
    
        pi(if_mod#) = []
        pi(true) = []
        pi(s) = []
        pi(mod#) = []
        pi(minus) = []
        pi(le) = []
        pi(|0|) = []
        pi(false) = []
        pi(pred) = []
        pi(mod) = []
        pi(if_mod) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            if_mod#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + ((0,1),(0,1)) x3 + (1,1)
            true_A() = (9,1)
            s_A(x1) = ((1,1),(1,1)) x1 + (20,4)
            mod#_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,1),(0,1)) x2 + (1,1)
            minus_A(x1,x2) = x1 + (1,2)
            le_A(x1,x2) = (10,1)
            |0|_A() = (0,1)
            false_A() = (20,5)
            pred_A(x1) = x1
            mod_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(0,1)) x2 + (19,0)
            if_mod_A(x1,x2,x3) = ((0,1),(0,1)) x2 + ((0,0),(0,1)) x3 + (18,0)
            rand_A(x1) = (21,4)
    
      precedence:
      
        mod# > le = mod = if_mod > |0| > false > s = minus > pred > if_mod# = true = rand
      
      partial status:
    
        pi(if_mod#) = []
        pi(true) = []
        pi(s) = []
        pi(mod#) = []
        pi(minus) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(false) = []
        pi(pred) = []
        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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)
r12: rand(x) -> x
r13: 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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)
r12: rand(x) -> x
r13: 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

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{11, x1 + 1}
          s_A(x1) = max{10, x1}
          le_A(x1,x2) = x1 + 11
          |0|_A = 4
          true_A = 3
          false_A = 0
          pred_A(x1) = max{1, x1}
          minus_A(x1,x2) = max{3, x1, x2 + 2}
          mod_A(x1,x2) = max{x1 + 5, x2 + 9}
          if_mod_A(x1,x2,x3) = max{x1 - 2, x2 + 5, x3 + 9}
          rand_A(x1) = max{21, x1 + 11}
    
      precedence:
      
        minus > le# = pred = mod = if_mod > |0| > s = le = true > false = rand
      
      partial status:
    
        pi(le#) = [1]
        pi(s) = [1]
        pi(le) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(pred) = [1]
        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{0, x1 - 1}
          s_A(x1) = x1
          le_A(x1,x2) = x1 + 5
          |0|_A = 0
          true_A = 0
          false_A = 2
          pred_A(x1) = max{5, x1}
          minus_A(x1,x2) = max{x1 + 5, x2 + 5}
          mod_A(x1,x2) = 0
          if_mod_A(x1,x2,x3) = 0
          rand_A(x1) = 1
    
      precedence:
      
        false = minus > |0| = true = pred > le# = s = le = mod = if_mod = rand
      
      partial status:
    
        pi(le#) = []
        pi(s) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(pred) = [1]
        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#(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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)
r12: rand(x) -> x
r13: 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

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, x2 + 1}
          s_A(x1) = max{2, x1}
          le_A(x1,x2) = max{x1 + 7, x2 + 4}
          |0|_A = 12
          true_A = 11
          false_A = 8
          pred_A(x1) = x1
          minus_A(x1,x2) = max{x1, x2 + 1}
          mod_A(x1,x2) = max{8, x1 + 3, x2 + 6}
          if_mod_A(x1,x2,x3) = max{8, x2 + 3, x3 + 6}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        minus# = false > minus > true = pred = mod = if_mod > |0| > s = le > rand
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(pred) = [1]
        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{19, x2}
          s_A(x1) = max{19, x1 + 7}
          le_A(x1,x2) = max{x1 - 8, x2 + 6}
          |0|_A = 1
          true_A = 2
          false_A = 6
          pred_A(x1) = max{23, x1}
          minus_A(x1,x2) = max{x1 + 24, x2 + 17}
          mod_A(x1,x2) = 0
          if_mod_A(x1,x2,x3) = 0
          rand_A(x1) = 1
    
      precedence:
      
        mod = if_mod > minus# = s = le = |0| = true = false = pred = minus = rand
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
        pi(le) = [2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(pred) = []
        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.