YES

We show the termination of the TRS R:

  min(X,|0|()) -> X
  min(s(X),s(Y)) -> min(X,Y)
  quot(|0|(),s(Y)) -> |0|()
  quot(s(X),s(Y)) -> s(quot(min(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: min#(s(X),s(Y)) -> min#(X,Y)
p2: quot#(s(X),s(Y)) -> quot#(min(X,Y),s(Y))
p3: quot#(s(X),s(Y)) -> min#(X,Y)
p4: log#(s(s(X))) -> log#(s(quot(X,s(s(|0|())))))
p5: log#(s(s(X))) -> quot#(X,s(s(|0|())))

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p2}
  {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: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        min > |0| > quot > s > log#
      
      argument filter:
    
        pi(log#) = [1]
        pi(s) = 1
        pi(quot) = 1
        pi(|0|) = []
        pi(min) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        min > |0| > quot > log# > s
      
      argument filter:
    
        pi(log#) = [1]
        pi(s) = [1]
        pi(quot) = 1
        pi(|0|) = []
        pi(min) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |0| > quot > s > min > log#
      
      argument filter:
    
        pi(log#) = []
        pi(s) = [1]
        pi(quot) = [1]
        pi(|0|) = []
        pi(min) = 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#(min(X,Y),s(Y))

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > min > quot# > s
      
      argument filter:
    
        pi(quot#) = [1, 2]
        pi(s) = [1]
        pi(min) = 1
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > min > quot# > s
      
      argument filter:
    
        pi(quot#) = [1, 2]
        pi(s) = [1]
        pi(min) = [1]
        pi(|0|) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |0| > min > s > quot#
      
      argument filter:
    
        pi(quot#) = [1, 2]
        pi(s) = 1
        pi(min) = 1
        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: min#(s(X),s(Y)) -> min#(X,Y)

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: 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 > min#
      
      argument filter:
    
        pi(min#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > min#
      
      argument filter:
    
        pi(min#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > min#
      
      argument filter:
    
        pi(min#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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