YES

We show the termination of the TRS 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|()))))))

-- SCC decomposition.

Consider the 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|()))))))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

Consider the 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|()))))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          log#_A(x1) = max{21, x1 + 7}
          s_A(x1) = max{13, x1 + 11}
          quot_A(x1,x2) = x1 + 3
          |0|_A = 6
          le_A(x1,x2) = 6
          true_A = 1
          false_A = 5
          if_minus_A(x1,x2,x3) = max{10, x2}
          minus_A(x1,x2) = max{2, x1}
    
      precedence:
      
        quot = le = true > |0| = false = minus > log# = s = if_minus
      
      partial status:
    
        pi(log#) = [1]
        pi(s) = []
        pi(quot) = []
        pi(|0|) = []
        pi(le) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = [2]
        pi(minus) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          log#_A(x1) = 54
          s_A(x1) = 54
          quot_A(x1,x2) = 54
          |0|_A = 33
          le_A(x1,x2) = 74
          true_A = 35
          false_A = 55
          if_minus_A(x1,x2,x3) = max{32, x2 - 16}
          minus_A(x1,x2) = 53
    
      precedence:
      
        quot > log# = s > false = minus > |0| = le = true = if_minus
      
      partial status:
    
        pi(log#) = []
        pi(s) = []
        pi(quot) = []
        pi(|0|) = []
        pi(le) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = []
        pi(minus) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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|()))))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = x1 + 2
          s_A(x1) = x1 + 13
          minus_A(x1,x2) = x1 + 5
          le_A(x1,x2) = max{x1 + 8, x2 + 13}
          |0|_A = 11
          true_A = 12
          false_A = 12
          if_minus_A(x1,x2,x3) = max{9, x2 + 5}
    
      precedence:
      
        quot# = minus = |0| > le = true = false = if_minus > s
      
      partial status:
    
        pi(quot#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 9}
          s_A(x1) = max{9, x1 + 4}
          minus_A(x1,x2) = x1 + 4
          le_A(x1,x2) = max{x1 + 14, x2 - 5}
          |0|_A = 2
          true_A = 15
          false_A = 3
          if_minus_A(x1,x2,x3) = 0
    
      precedence:
      
        quot# = s = minus = le = |0| = true = false = if_minus
      
      partial status:
    
        pi(quot#) = []
        pi(s) = [1]
        pi(minus) = [1]
        pi(le) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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|()))))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{x1 + 2, x2 + 1, x3 + 4}
          false_A = 4
          s_A(x1) = x1 + 5
          minus#_A(x1,x2) = max{x1 + 6, x2 + 4}
          le_A(x1,x2) = max{x1 + 3, x2 - 6}
          |0|_A = 11
          true_A = 12
    
      precedence:
      
        if_minus# = s = minus# = le = |0| > false = true
      
      partial status:
    
        pi(if_minus#) = [2]
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{1, x2}
          false_A = 7
          s_A(x1) = x1 + 8
          minus#_A(x1,x2) = x1
          le_A(x1,x2) = 1
          |0|_A = 0
          true_A = 2
    
      precedence:
      
        true > if_minus# = false = s = minus# = le = |0|
      
      partial status:
    
        pi(if_minus#) = []
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

Consider the 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|()))))))

The set of usable rules consists of

  (no rules)

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{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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