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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        log#_A(x1) = max{5, x1 - 10}
        s_A(x1) = x1 + 9
        quot_A(x1,x2) = x1
        |0|_A = 4
        le_A(x1,x2) = max{8, x1 - 3}
        true_A = 5
        false_A = 7
        minus_A(x1,x2) = x1
        if_minus_A(x1,x2,x3) = max{8, x1 + 1, x2}
  
    base order 2:
  
      precedence:
      
        minus > true = if_minus > log# = s = |0| > quot = le > false
      
  
    partial status:
  
      pi(log#) = []
      pi(s) = []
      pi(quot) = [1]
      pi(|0|) = []
      pi(le) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = [1]
      pi(if_minus) = [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

  r1, r2, r4, r5, r6, r7, r8, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = max{3, x1 - 6}
        s_A(x1) = max{12, x1 + 10}
        minus_A(x1,x2) = max{5, x1 + 3}
        le_A(x1,x2) = max{1, x1 - 9, x2 - 13}
        |0|_A = 1
        true_A = 1
        false_A = 1
        if_minus_A(x1,x2,x3) = max{14, x2 + 3}
        quot_A(x1,x2) = max{24, x1 - 6, x2 - 33}
        log_A(x1) = max{23, x1 - 45}
  
    base order 2:
  
      precedence:
      
        le = quot > minus = true = if_minus = log > quot# = s > |0| > false
      
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(if_minus) = [2]
      pi(quot) = []
      pi(log) = []

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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_minus#_A(x1,x2,x3) = x2 + 7
        false_A = 7
        s_A(x1) = x1 + 10
        minus#_A(x1,x2) = x1 + 8
        le_A(x1,x2) = max{x1 + 2, x2 + 9}
        |0|_A = 8
        true_A = 1
        minus_A(x1,x2) = x2 + 22
        if_minus_A(x1,x2,x3) = max{22, x1 - 8, x2 - 15, x3 - 1}
        quot_A(x1,x2) = max{10, x1 - 72, x2 - 49}
        log_A(x1) = 29
  
    base order 2:
  
      precedence:
      
        log > quot > s = minus# = minus = if_minus > if_minus# = le > true > |0| > false
      
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = [1, 2]
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [2]
      pi(if_minus) = []
      pi(quot) = []
      pi(log) = []

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

  r1, r2, r4, r6, r8, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{x1 - 1, x2 + 8}
        s_A(x1) = max{27, x1 + 7}
        le_A(x1,x2) = max{17, x1 - 30, x2 - 28}
        |0|_A = 15
        true_A = 16
        false_A = 8
        minus_A(x1,x2) = max{18, x1 - 29, x2 - 15}
        if_minus_A(x1,x2,x3) = max{17, x1 - 18, x2 - 37, x3 - 9}
        quot_A(x1,x2) = max{16, x1 - 26, x2 - 47}
        log_A(x1) = max{35, x1 - 39}
  
    base order 2:
  
      precedence:
      
        minus > le > le# = s = true = if_minus > log > quot > |0| > false
      
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = []
      pi(quot) = []
      pi(log) = []

The following rules are removed:

  p1

Then no dependency pair remains.