YES

We show the termination of the TRS R:

  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(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: minus#(x,s(y)) -> pred#(minus(x,y))
p2: minus#(x,s(y)) -> minus#(x,y)
p3: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p4: quot#(s(x),s(y)) -> minus#(x,y)
p5: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|())))))
p6: log#(s(s(x))) -> quot#(x,s(s(|0|())))

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p3}
  {p2}


-- 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: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2, r3, r4, r5

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{9, x1 + 6}
          s_A(x1) = x1 + 14
          quot_A(x1,x2) = x1
          |0|_A = 8
          pred_A(x1) = max{0, x1 - 1}
          minus_A(x1,x2) = x1
    
      precedence:
      
        |0| = pred = minus > log# = quot > s
      
      partial status:
    
        pi(log#) = [1]
        pi(s) = [1]
        pi(quot) = [1]
        pi(|0|) = []
        pi(pred) = []
        pi(minus) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          log#_A(x1) = 1
          s_A(x1) = max{5, x1}
          quot_A(x1,x2) = max{12, x1 + 7}
          |0|_A = 15
          pred_A(x1) = 4
          minus_A(x1,x2) = 5
    
      precedence:
      
        pred > quot > |0| > s > log# = minus
      
      partial status:
    
        pi(log#) = []
        pi(s) = []
        pi(quot) = []
        pi(|0|) = []
        pi(pred) = []
        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: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: 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:
        
          quot#_A(x1,x2) = max{4, x1 + 1}
          s_A(x1) = x1 + 6
          minus_A(x1,x2) = max{2, x1 + 1}
          pred_A(x1) = x1
          |0|_A = 0
    
      precedence:
      
        quot# = s = minus = pred = |0|
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        pi(pred) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = 0
          s_A(x1) = x1 + 3
          minus_A(x1,x2) = 1
          pred_A(x1) = max{1, x1 - 1}
          |0|_A = 0
    
      precedence:
      
        minus = pred > quot# = s = |0|
      
      partial status:
    
        pi(quot#) = []
        pi(s) = [1]
        pi(minus) = []
        pi(pred) = []
        pi(|0|) = []
    

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: minus#(x,s(y)) -> minus#(x,y)

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: 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:
        
          minus#_A(x1,x2) = max{2, x1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        minus# = s
      
      partial status:
    
        pi(minus#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{x1, x2 - 1}
          s_A(x1) = x1
    
      precedence:
      
        minus# = s
      
      partial status:
    
        pi(minus#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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