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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          log#_A(x1) = x1 + 1
          s_A(x1) = x1 + 3
          quot_A(x1,x2) = x1
          |0|_A() = 2
          le_A(x1,x2) = x1 + 2
          true_A() = 3
          false_A() = 1
          minus_A(x1,x2) = x1
          if_minus_A(x1,x2,x3) = x2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          log#_A(x1) = 0
          s_A(x1) = max{2, x1 - 7}
          quot_A(x1,x2) = max{64, x1 + 35, x2 - 10}
          |0|_A = 29
          le_A(x1,x2) = 90
          true_A = 90
          false_A = 63
          minus_A(x1,x2) = max{29, x1 - 30, x2 - 91}
          if_minus_A(x1,x2,x3) = max{29, x1 - 91, x2 - 30, x3 - 91}
          log#!_A(x1) = x1 + 19
          s!_A(x1) = max{49, x1 - 4}
          quot!_A(x1,x2) = max{x1 + 58, x2 - 14}
          |0|!_A = 88
          le!_A(x1,x2) = max{91, x1 - 29}
          true!_A = 89
          false!_A = 1
          minus!_A(x1,x2) = max{x1 - 3, x2 + 90}
          if_minus!_A(x1,x2,x3) = max{x1 - 1, x2 - 7, x3 + 89}
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = []
      pi(quot) = [1]
      pi(|0|) = []
      pi(le) = [1]
      pi(true) = []
      pi(false) = []
      pi(minus) = [1]
      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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          quot#_A(x1,x2) = x1 + 1
          s_A(x1) = x1 + 5
          minus_A(x1,x2) = x1 + 3
          |0|_A() = 7
          if_minus_A(x1,x2,x3) = x2 + 3
          le_A(x1,x2) = 4
          true_A() = 8
          false_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 37, x2 - 9}
          s_A(x1) = max{8, x1 - 14}
          minus_A(x1,x2) = x2 + 14
          |0|_A = 7
          if_minus_A(x1,x2,x3) = max{8, x3 + 6}
          le_A(x1,x2) = max{23, x1 + 15, x2 - 1}
          true_A = 26
          false_A = 24
          quot#!_A(x1,x2) = x2 + 23
          s!_A(x1) = 32
          minus!_A(x1,x2) = max{x1 + 45, x2 + 8}
          |0|!_A = 32
          if_minus!_A(x1,x2,x3) = x2 + 25
          le!_A(x1,x2) = 0
          true!_A = 1
          false!_A = 31
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(if_minus) = [2]
      pi(le) = []
      pi(true) = []
      pi(false) = []

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

  r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_minus#_A(x1,x2,x3) = 4
          false_A() = 9
          s_A(x1) = 5
          minus#_A(x1,x2) = 4
          le_A(x1,x2) = x1 + x2 + 1
          log_A(x1) = 3
          |0|_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{1, x2 - 11, x3 - 2}
          false_A = 1
          s_A(x1) = max{10, x1 + 3}
          minus#_A(x1,x2) = max{1, x1 - 11, x2 - 2}
          le_A(x1,x2) = max{0, x1 - 14, x2 - 11}
          log_A(x1) = max{11, x1 - 27}
          |0|_A = 10
          if_minus#!_A(x1,x2,x3) = max{1, x2 - 7}
          false!_A = 0
          s!_A(x1) = 0
          minus#!_A(x1,x2) = max{2, x1 - 7}
          le!_A(x1,x2) = max{0, x1 - 13, x2 - 11}
          log!_A(x1) = max{1, x1 - 18}
          |0|!_A = 0
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = []
      pi(minus#) = []
      pi(le) = []
      pi(log) = []
      pi(|0|) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 6
  
    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{6, x1 - 4}
          le#!_A(x1,x2) = max{1, x1 - 7, x2 - 7}
          s!_A(x1) = max{6, x1 - 28}
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.