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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = max{0, x1 - 34}
            s_A(x1) = x1 + 33
            quot_A(x1,x2) = x1 + 21
            |0|_A = 8
            le_A(x1,x2) = x2 + 39
            true_A = 36
            false_A = 1
            minus_A(x1,x2) = x1
            if_minus_A(x1,x2,x3) = max{11, x2}
            log#!_A(x1) = max{3, x1 - 10}
            s!_A(x1) = max{23, x1 + 10}
            quot!_A(x1,x2) = x1 + 22
            |0|!_A = 2
            le!_A(x1,x2) = max{5, x2 - 1}
            true!_A = 1
            false!_A = 6
            minus!_A(x1,x2) = max{4, x1 - 9}
            if_minus!_A(x1,x2,x3) = max{3, x2 - 23}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = 0
            s_A(x1) = 28
            quot_A(x1,x2) = max{24, x1 + 21, x2 - 23}
            |0|_A = 6
            le_A(x1,x2) = 22
            true_A = 22
            false_A = 21
            minus_A(x1,x2) = max{1, x1}
            if_minus_A(x1,x2,x3) = max{28, x1 + 6, x2 - 19}
            log#!_A(x1) = max{0, x1 - 44}
            s!_A(x1) = max{0, x1 - 2}
            quot!_A(x1,x2) = max{6, x1 - 19, x2 - 45}
            |0|!_A = 27
            le!_A(x1,x2) = 0
            true!_A = 1
            false!_A = 28
            minus!_A(x1,x2) = x2 + 26
            if_minus!_A(x1,x2,x3) = max{9, x1 - 10, x2 - 20}
        
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = []
      pi(quot) = [1]
      pi(|0|) = []
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = [2]

The following rules are removed:

  p1

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, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 35, x2 - 29}
            s_A(x1) = max{28, x1 + 18}
            minus_A(x1,x2) = max{15, x1 + 5}
            le_A(x1,x2) = 11
            |0|_A = 15
            true_A = 1
            false_A = 11
            if_minus_A(x1,x2,x3) = max{27, x1 + 13, x2 + 5}
            quot_A(x1,x2) = max{39, x1 - 34, x2 - 4}
            quot#!_A(x1,x2) = max{1, x1 - 16}
            s!_A(x1) = max{11, x1 - 23}
            minus!_A(x1,x2) = max{62, x1 + 42}
            le!_A(x1,x2) = 61
            |0|!_A = 10
            true!_A = 11
            false!_A = 61
            if_minus!_A(x1,x2,x3) = max{61, x2 - 36}
            quot!_A(x1,x2) = max{28, x1 + 25, x2 + 11}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 22, x2 - 121}
            s_A(x1) = x1 + 21
            minus_A(x1,x2) = x1
            le_A(x1,x2) = max{2, x1 - 10}
            |0|_A = 11
            true_A = 2
            false_A = 6
            if_minus_A(x1,x2,x3) = max{10, x1 + 2, x2}
            quot_A(x1,x2) = max{22, x1 - 22, x2 - 100}
            quot#!_A(x1,x2) = x1 + 77
            s!_A(x1) = 4
            minus!_A(x1,x2) = max{x1 + 99, x2 + 2}
            le!_A(x1,x2) = 119
            |0|!_A = 3
            true!_A = 119
            false!_A = 119
            if_minus!_A(x1,x2,x3) = x1 + 1
            quot!_A(x1,x2) = max{6, x1 - 22, x2 - 89}
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(if_minus) = []
      pi(quot) = [2]

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_minus#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 37, x3 - 4}
            false_A = 0
            s_A(x1) = max{34, x1 + 25}
            minus#_A(x1,x2) = max{0, x1 - 37, x2 - 4}
            le_A(x1,x2) = max{0, x1 - 36, x2 - 21}
            |0|_A = 20
            true_A = 0
            if_minus#!_A(x1,x2,x3) = max{2, x1 - 45, x2 - 27, x3 - 1}
            false!_A = 1
            s!_A(x1) = max{3, x1 - 46}
            minus#!_A(x1,x2) = max{7, x1 - 26, x2 - 1}
            le!_A(x1,x2) = max{6, x1 - 110, x2 - 44}
            |0|!_A = 2
            true!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_minus#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 45, x3 - 16}
            false_A = 0
            s_A(x1) = max{42, x1 + 36}
            minus#_A(x1,x2) = max{0, x1 - 45, x2 - 16}
            le_A(x1,x2) = max{0, x1 - 44, x2 - 15}
            |0|_A = 14
            true_A = 0
            if_minus#!_A(x1,x2,x3) = max{9, x1 - 42, x2 - 84, x3 - 43}
            false!_A = 4
            s!_A(x1) = max{8, x1 - 51}
            minus#!_A(x1,x2) = max{9, x1 - 83, x2 - 43}
            le!_A(x1,x2) = max{10, x1 - 88, x2 - 44}
            |0|!_A = 3
            true!_A = 2
        
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{3, x1 - 5}
            le#!_A(x1,x2) = max{0, x2 - 75}
            s!_A(x1) = max{5, x1 - 74}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{9, x1 - 11}
            le#!_A(x1,x2) = max{6, x1 - 168, x2 - 10}
            s!_A(x1) = max{5, x1 - 168}
        
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.