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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        log#_A(x1) = ((1,1),(0,0)) x1
        s_A(x1) = x1 + (3,0)
        quot_A(x1,x2) = x1 + (1,1)
        |0|_A() = (1,1)
        min_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        log#_A(x1) = (0,0)
        s_A(x1) = (1,1)
        quot_A(x1,x2) = (2,1)
        |0|_A() = (1,2)
        min_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        log#_A(x1) = (0,0)
        s_A(x1) = (3,2)
        quot_A(x1,x2) = (2,1)
        |0|_A() = (1,1)
        min_A(x1,x2) = (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,0),(1,0)) x1
        s_A(x1) = ((1,1),(0,1)) x1 + (2,1)
        min_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,0),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        min_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,0),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        min_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
    

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: 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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = ((1,0),(1,1)) x1 + x2
        s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

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