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. lexicographic path order with precedence:
    
      precedence:
      
        log# > quot > s > |0| > if_minus > le > minus > false > true
      
      argument filter:
    
        pi(log#) = 1
        pi(s) = [1]
        pi(quot) = 1
        pi(|0|) = []
        pi(le) = [1]
        pi(true) = []
        pi(false) = []
        pi(if_minus) = 2
        pi(minus) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        le > minus > quot > |0| > if_minus > true > log# > s > false
      
      argument filter:
    
        pi(log#) = []
        pi(s) = []
        pi(quot) = [1]
        pi(|0|) = []
        pi(le) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = [2]
        pi(minus) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |0| > if_minus > le > minus > false > true > s > quot > log#
      
      argument filter:
    
        pi(log#) = []
        pi(s) = []
        pi(quot) = []
        pi(|0|) = []
        pi(le) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = 2
        pi(minus) = [1]
    

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. lexicographic path order with precedence:
    
      precedence:
      
        s > minus > le > true > |0| > if_minus > false > quot#
      
      argument filter:
    
        pi(quot#) = 1
        pi(s) = [1]
        pi(minus) = 1
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = 2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > minus > |0| > if_minus > false > true > le > quot#
      
      argument filter:
    
        pi(quot#) = 1
        pi(s) = [1]
        pi(minus) = 1
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = 2
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        quot# > s > minus > if_minus > |0| > false > le > true
      
      argument filter:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = 2
    

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. lexicographic path order with precedence:
    
      precedence:
      
        le > true > false > |0| > if_minus# > s > minus#
      
      argument filter:
    
        pi(if_minus#) = 2
        pi(false) = []
        pi(s) = 1
        pi(minus#) = 1
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        le > true > false > |0| > minus# > if_minus# > s
      
      argument filter:
    
        pi(if_minus#) = 2
        pi(false) = []
        pi(s) = 1
        pi(minus#) = 1
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        true > false > |0| > le > if_minus# > s > minus#
      
      argument filter:
    
        pi(if_minus#) = 2
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = 2
        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. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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