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))
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  log(s(|0|())) -> |0|()
  log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

  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: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p6: quot#(s(x),s(y)) -> minus#(x,y)
p7: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|())))))
p8: log#(s(s(x))) -> quot#(x,s(s(|0|())))

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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
r12: rand(x) -> x
r13: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|())))))

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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          log#_A(x1) = x1
          s_A(x1) = x1 + (0,4)
          quot_A(x1,x2) = x1
          |0|_A() = (0,1)
          le_A(x1,x2) = x2 + (2,3)
          true_A() = (1,2)
          false_A() = (1,2)
          minus_A(x1,x2) = x1
          if_minus_A(x1,x2,x3) = x2
          log_A(x1) = x1
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      le = true = false = rand > minus > log# = s = quot = |0| = if_minus = log
    
    partial status:
  
      pi(log#) = []
      pi(s) = []
      pi(quot) = []
      pi(|0|) = []
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = []
      pi(log) = [1]
      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: quot#(s(x),s(y)) -> quot#(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(|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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          quot#_A(x1,x2) = x1
          s_A(x1) = x1 + (0,3)
          minus_A(x1,x2) = x1
          le_A(x1,x2) = x1 + (2,2)
          |0|_A() = (0,2)
          true_A() = (1,3)
          false_A() = (1,4)
          if_minus_A(x1,x2,x3) = x2
          quot_A(x1,x2) = x1
          log_A(x1) = x1 + (1,1)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
  
    precedence:
    
      s = if_minus = quot = log > |0| = false = rand > quot# = minus = le = true
    
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(if_minus) = []
      pi(quot) = []
      pi(log) = [1]
      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: 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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          if_minus#_A(x1,x2,x3) = ((1,0),(1,1)) x1 + x2 + ((1,0),(1,1)) x3 + (1,0)
          false_A() = (2,4)
          s_A(x1) = x1 + (0,3)
          minus#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (3,8)
          le_A(x1,x2) = (2,5)
          |0|_A() = (0,2)
          true_A() = (1,0)
          minus_A(x1,x2) = x1
          if_minus_A(x1,x2,x3) = x2
          quot_A(x1,x2) = x1
          log_A(x1) = x1 + (1,4)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,4)
  
    precedence:
    
      rand > log > |0| = minus = if_minus = quot > if_minus# = s = minus# = le > false > true
    
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = []
      pi(if_minus) = []
      pi(quot) = [1]
      pi(log) = [1]
      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: 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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
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: 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: quot(|0|(),s(y)) -> |0|()
r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r10: log(s(|0|())) -> |0|()
r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          le#_A(x1,x2) = x1
          s_A(x1) = x1 + (0,6)
          le_A(x1,x2) = x2 + (2,5)
          |0|_A() = (0,3)
          true_A() = (1,4)
          false_A() = (1,7)
          minus_A(x1,x2) = x1
          if_minus_A(x1,x2,x3) = x2
          quot_A(x1,x2) = x1
          log_A(x1) = x1 + (1,7)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      le = false > |0| = true = minus = log = rand > le# = if_minus = quot > s
    
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = [2]
      pi(quot) = [1]
      pi(log) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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