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:
  
      max/plus interpretation on natural numbers:
      
        log#_A(x1) = max{12, x1}
        s_A(x1) = x1 + 6
        quot_A(x1,x2) = x1 + 4
        |0|_A = 4
        le_A(x1,x2) = max{x1 + 7, x2 + 15}
        true_A = 14
        false_A = 14
        minus_A(x1,x2) = x1
        if_minus_A(x1,x2,x3) = max{3, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          log#_A(x1) = max{0, x1 - 10}
          s_A(x1) = max{9, x1 - 26}
          quot_A(x1,x2) = max{25, x1 - 40, x2 - 51}
          |0|_A = 2
          le_A(x1,x2) = 34
          true_A = 34
          false_A = 33
          minus_A(x1,x2) = max{25, x1 - 33, x2 - 33}
          if_minus_A(x1,x2,x3) = max{9, x1 - 35, x2 - 33, x3 - 33}
          log#!_A(x1) = 1
          s!_A(x1) = 3
          quot!_A(x1,x2) = x1 + 32
          |0|!_A = 66
          le!_A(x1,x2) = max{75, x1 - 3}
          true!_A = 67
          false!_A = 4
          minus!_A(x1,x2) = max{x1 + 65, x2 - 2}
          if_minus!_A(x1,x2,x3) = max{4, x1 - 35, x2 - 10, x3 - 3}
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = [1]
      pi(quot) = [1]
      pi(|0|) = []
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = []

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

  r4, r5, r6, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = 5
        s_A(x1) = x1 + 10
        minus_A(x1,x2) = x1 + 9
        |0|_A = 11
        if_minus_A(x1,x2,x3) = max{18, x2 + 9}
        le_A(x1,x2) = 3
        true_A = 2
        false_A = 2
        quot_A(x1,x2) = max{12, x1 - 12, x2 - 48}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{0, x2 - 39}
          s_A(x1) = max{61, x1 + 38}
          minus_A(x1,x2) = max{24, x1 + 12}
          |0|_A = 2
          if_minus_A(x1,x2,x3) = x2 + 12
          le_A(x1,x2) = max{58, x1 - 3, x2 + 56}
          true_A = 61
          false_A = 59
          quot_A(x1,x2) = max{70, x1 + 46, x2 - 69}
          quot#!_A(x1,x2) = max{1, x1 - 25}
          s!_A(x1) = max{21, x1 - 14}
          minus!_A(x1,x2) = max{49, x1 - 15}
          |0|!_A = 48
          if_minus!_A(x1,x2,x3) = max{47, x2 - 39}
          le!_A(x1,x2) = 0
          true!_A = 1
          false!_A = 47
          quot!_A(x1,x2) = 0
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(if_minus) = [2]
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(quot) = []

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, r4, r6, r8, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_minus#_A(x1,x2,x3) = max{x1 - 47, x2 - 27, x3 + 8}
        false_A = 17
        s_A(x1) = max{18, x1 + 7}
        minus#_A(x1,x2) = max{x1 - 26, x2 + 8}
        le_A(x1,x2) = max{27, x1 - 25, x2 - 25}
        |0|_A = 24
        true_A = 25
        minus_A(x1,x2) = max{26, x2 - 9}
        if_minus_A(x1,x2,x3) = max{34, x1 - 28, x2 - 19, x3 - 1}
        quot_A(x1,x2) = max{26, x1 - 36, x2 - 53}
        log_A(x1) = max{39, x1 - 41}
  
    base order 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 - 19, x3 - 6}
          false_A = 0
          s_A(x1) = max{17, x1 + 14}
          minus#_A(x1,x2) = max{0, x1 - 19, x2 - 6}
          le_A(x1,x2) = max{0, x1 - 18, x2 - 5}
          |0|_A = 0
          true_A = 0
          minus_A(x1,x2) = max{15, x1 - 20, x2 - 7}
          if_minus_A(x1,x2,x3) = max{16, x1 - 1, x2 - 21, x3 - 1}
          quot_A(x1,x2) = max{15, x1 - 30, x2 - 38}
          log_A(x1) = max{16, x1 - 44}
          if_minus#!_A(x1,x2,x3) = max{8, x2 - 13}
          false!_A = 6
          s!_A(x1) = 5
          minus#!_A(x1,x2) = max{7, x1 - 5}
          le!_A(x1,x2) = max{13, x1 - 18, x2 - 18}
          |0|!_A = 1
          true!_A = 2
          minus!_A(x1,x2) = max{2, x1 - 20, x2 - 48}
          if_minus!_A(x1,x2,x3) = max{6, x1 - 1, x2 - 18, x3 - 1}
          quot!_A(x1,x2) = max{5, x1 - 18, x2 - 61}
          log!_A(x1) = max{6, x1 - 46}
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = []
      pi(minus#) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = []
      pi(if_minus) = []
      pi(quot) = []
      pi(log) = []

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

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{x1 - 1, x2 + 5}
        s_A(x1) = max{8, x1 + 4}
        le_A(x1,x2) = max{3, x1 - 8, x2 - 5}
        |0|_A = 0
        true_A = 1
        false_A = 3
        minus_A(x1,x2) = max{4, x1}
        if_minus_A(x1,x2,x3) = max{7, x2}
        quot_A(x1,x2) = max{3, x1 + 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{7, x1 - 13}
          le_A(x1,x2) = 25
          |0|_A = 3
          true_A = 20
          false_A = 25
          minus_A(x1,x2) = max{12, x1 - 4, x2 + 1}
          if_minus_A(x1,x2,x3) = max{2, x1 - 18, x2 - 4, x3 - 2}
          quot_A(x1,x2) = max{8, x1 - 45, x2 - 21}
          le#!_A(x1,x2) = 0
          s!_A(x1) = max{1, x1 - 18}
          le!_A(x1,x2) = 0
          |0|!_A = 0
          true!_A = 0
          false!_A = 0
          minus!_A(x1,x2) = max{0, x1 - 5, x2 - 17}
          if_minus!_A(x1,x2,x3) = max{2, x1 - 44, x2 - 8, x3 - 17}
          quot!_A(x1,x2) = max{3, x1 - 8, x2 - 11}
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = []
      pi(quot) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.