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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            log#_A(x1) = x1
            s_A(x1) = x1 + (0,3)
            quot_A(x1,x2) = x1
            |0|_A() = (0,2)
            le_A(x1,x2) = (2,5)
            true_A() = (1,1)
            false_A() = (1,4)
            if_minus_A(x1,x2,x3) = x2
            minus_A(x1,x2) = x1
    
      precedence:
      
        log# > le = true = if_minus = minus > s = quot = |0| = false
      
      partial status:
    
        pi(log#) = [1]
        pi(s) = [1]
        pi(quot) = [1]
        pi(|0|) = []
        pi(le) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = []
        pi(minus) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            log#_A(x1) = x1 + (3,4)
            s_A(x1) = ((1,0),(1,1)) x1 + (5,4)
            quot_A(x1,x2) = x1 + (1,3)
            |0|_A() = (2,14)
            le_A(x1,x2) = (4,7)
            true_A() = (1,0)
            false_A() = (1,5)
            if_minus_A(x1,x2,x3) = (6,15)
            minus_A(x1,x2) = (0,6)
    
      precedence:
      
        quot = if_minus = minus > |0| = false > s > log# > le = true
      
      partial status:
    
        pi(log#) = [1]
        pi(s) = [1]
        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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = x1 + (1,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (5,5)
            minus_A(x1,x2) = ((1,0),(1,0)) x1 + (4,3)
            le_A(x1,x2) = ((0,0),(1,0)) x1 + (3,2)
            |0|_A() = (1,1)
            true_A() = (2,2)
            false_A() = (2,6)
            if_minus_A(x1,x2,x3) = x2 + (4,1)
    
      precedence:
      
        minus > le > |0| > true > quot# = s = false = if_minus
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(if_minus) = [2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = (0,1)
            s_A(x1) = (2,2)
            minus_A(x1,x2) = (4,3)
            le_A(x1,x2) = (1,4)
            |0|_A() = (0,0)
            true_A() = (0,0)
            false_A() = (0,0)
            if_minus_A(x1,x2,x3) = (3,0)
    
      precedence:
      
        if_minus > quot# = s = minus = le = |0| = true = false
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        pi(le) = []
        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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_minus#_A(x1,x2,x3) = x2 + ((0,0),(1,0)) x3 + (1,1)
            false_A() = (4,1)
            s_A(x1) = ((1,0),(1,0)) x1 + (3,2)
            minus#_A(x1,x2) = ((1,0),(0,0)) x1 + (2,0)
            le_A(x1,x2) = x2 + (2,1)
            |0|_A() = (5,2)
            true_A() = (2,1)
    
      precedence:
      
        false > le = true > |0| > if_minus# = s = minus#
      
      partial status:
    
        pi(if_minus#) = [2]
        pi(false) = []
        pi(s) = []
        pi(minus#) = []
        pi(le) = [2]
        pi(|0|) = []
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if_minus#_A(x1,x2,x3) = (0,1)
            false_A() = (2,1)
            s_A(x1) = (1,0)
            minus#_A(x1,x2) = (0,0)
            le_A(x1,x2) = ((1,0),(1,1)) x2 + (2,1)
            |0|_A() = (3,2)
            true_A() = (0,0)
    
      precedence:
      
        le > if_minus# = s > minus# > false = |0| > true
      
      partial status:
    
        pi(if_minus#) = []
        pi(false) = []
        pi(s) = []
        pi(minus#) = []
        pi(le) = [2]
        pi(|0|) = []
        pi(true) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: if_minus#(false(),s(x),y) -> minus#(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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            le#_A(x1,x2) = ((1,0),(0,0)) x1
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        le# > s
      
      partial status:
    
        pi(le#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            le#_A(x1,x2) = (0,0)
            s_A(x1) = (1,1)
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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