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, r10

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              log#_A(x1) = x1
              s_A(x1) = x1 + 4
              quot_A(x1,x2) = x1
              |0|_A() = 0
              le_A(x1,x2) = 1
              true_A() = 1
              false_A() = 1
              minus_A(x1,x2) = x1
              if_minus_A(x1,x2,x3) = x2
              log_A(x1) = 0
              log#!_A(x1) = x1
              s!_A(x1) = 9
              quot!_A(x1,x2) = x2
              |0|!_A() = 5
              le!_A(x1,x2) = 8
              true!_A() = 0
              false!_A() = 7
              minus!_A(x1,x2) = x1 + 6
              if_minus!_A(x1,x2,x3) = x2 + 5
              log!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = max{22, x1 + 1}
            s_A(x1) = max{10, x1}
            quot_A(x1,x2) = max{21, x1}
            |0|_A = 0
            le_A(x1,x2) = 25
            true_A = 13
            false_A = 25
            minus_A(x1,x2) = 11
            if_minus_A(x1,x2,x3) = max{9, x1 - 14}
            log_A(x1) = max{11, x1 - 21}
            log#!_A(x1) = max{1, x1 - 8}
            s!_A(x1) = 0
            quot!_A(x1,x2) = max{0, x1 - 7, x2 - 22}
            |0|!_A = 3
            le!_A(x1,x2) = 0
            true!_A = 4
            false!_A = 6
            minus!_A(x1,x2) = max{x1 - 1, x2 + 5}
            if_minus!_A(x1,x2,x3) = max{4, x2 - 9}
            log!_A(x1) = max{11, x1 - 11}
        
  
    partial status:
  
      pi(log#) = []
      pi(s) = []
      pi(quot) = [2]
      pi(|0|) = []
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = [1, 2]
      pi(if_minus) = []
      pi(log) = [1]

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

  r2, r4, r5, r6, r7

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              quot#_A(x1,x2) = 0
              s_A(x1) = 2
              minus_A(x1,x2) = x2 + 3
              le_A(x1,x2) = x2
              |0|_A() = 1
              false_A() = 1
              if_minus_A(x1,x2,x3) = x3 + 3
              true_A() = 1
              quot#!_A(x1,x2) = 5
              s!_A(x1) = 6
              minus!_A(x1,x2) = 1
              le!_A(x1,x2) = x2 + 5
              |0|!_A() = 3
              false!_A() = 2
              if_minus!_A(x1,x2,x3) = 0
              true!_A() = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 11, x2 - 58}
            s_A(x1) = max{53, x1 + 10}
            minus_A(x1,x2) = max{7, x1}
            le_A(x1,x2) = max{45, x1 - 9, x2 - 1}
            |0|_A = 5
            false_A = 45
            if_minus_A(x1,x2,x3) = max{16, x2}
            true_A = 45
            quot#!_A(x1,x2) = max{1, x1 - 8}
            s!_A(x1) = max{0, x1 - 12}
            minus!_A(x1,x2) = max{4, x1 + 3, x2 + 2}
            le!_A(x1,x2) = max{x1 + 1, x2 + 1}
            |0|!_A = 0
            false!_A = 5
            if_minus!_A(x1,x2,x3) = max{0, x2 - 50}
            true!_A = 6
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(le) = [1, 2]
      pi(|0|) = []
      pi(false) = []
      pi(if_minus) = [2]
      pi(true) = []

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, r4, r6, r8, r10

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_minus#_A(x1,x2,x3) = x2
              false_A() = 4
              s_A(x1) = x1 + 2
              minus#_A(x1,x2) = x1
              le_A(x1,x2) = x1 + 1
              |0|_A() = 0
              true_A() = 1
              minus_A(x1,x2) = x2 + 3
              if_minus_A(x1,x2,x3) = 4
              quot_A(x1,x2) = 3
              log_A(x1) = x1
              if_minus#!_A(x1,x2,x3) = 2
              false!_A() = 4
              s!_A(x1) = 7
              minus#!_A(x1,x2) = 3
              le!_A(x1,x2) = 2
              |0|!_A() = 0
              true!_A() = 1
              minus!_A(x1,x2) = x2 + 5
              if_minus!_A(x1,x2,x3) = 7
              quot!_A(x1,x2) = x2 + 4
              log!_A(x1) = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_minus#_A(x1,x2,x3) = max{1, x2 - 17, x3 - 2}
            false_A = 0
            s_A(x1) = max{16, x1 + 6}
            minus#_A(x1,x2) = max{1, x1 - 16, x2 - 2}
            le_A(x1,x2) = max{13, x1 - 16, x2 - 10}
            |0|_A = 0
            true_A = 0
            minus_A(x1,x2) = max{11, x1 - 18, x2 - 4}
            if_minus_A(x1,x2,x3) = max{15, x1 - 1, x2 - 19, x3 - 1}
            quot_A(x1,x2) = max{11, x1 - 31, x2 - 36}
            log_A(x1) = max{7, x1 - 24}
            if_minus#!_A(x1,x2,x3) = 1
            false!_A = 3
            s!_A(x1) = 2
            minus#!_A(x1,x2) = max{2, x1 - 17, x2 - 1}
            le!_A(x1,x2) = 2
            |0|!_A = 40
            true!_A = 41
            minus!_A(x1,x2) = max{2, x1 - 29, x2 - 40}
            if_minus!_A(x1,x2,x3) = max{2, x1 - 14, x2 - 17}
            quot!_A(x1,x2) = max{0, x1 - 36, x2 - 47}
            log!_A(x1) = max{41, x1 + 17}
        
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [2]
      pi(if_minus) = [1, 3]
      pi(quot) = [1, 2]
      pi(log) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- 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

  r4, r8, r10

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              minus_A(x1,x2) = x1 + 2
              |0|_A() = 3
              quot_A(x1,x2) = x1 + 2
              log_A(x1) = x1
              le#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 2
              minus!_A(x1,x2) = 5
              |0|!_A() = 1
              quot!_A(x1,x2) = 4
              log!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{23, x1 - 27}
            minus_A(x1,x2) = max{13, x1 - 132, x2 - 132}
            |0|_A = 11
            quot_A(x1,x2) = max{18, x1 - 33, x2 + 17}
            log_A(x1) = max{101, x1 - 22}
            le#!_A(x1,x2) = max{43, x2 - 24}
            s!_A(x1) = max{42, x1 + 30}
            minus!_A(x1,x2) = max{57, x1 - 23, x2 - 25}
            |0|!_A = 2
            quot!_A(x1,x2) = max{56, x2 + 33}
            log!_A(x1) = x1 + 109
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(quot) = []
      pi(log) = []

The following rules are removed:

  p1

Then no dependency pair remains.