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

  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{10, x1 - 12}
        s_A(x1) = 11
        quot_A(x1,x2) = max{x1 - 1, x2 + 15}
        |0|_A = 9
        minus_A(x1,x2) = max{x1 + 4, x2 + 13}
        if_minus_A(x1,x2,x3) = max{12, x2 - 1}
        le_A(x1,x2) = max{14, x1 - 10, x2 + 1}
        true_A = 10
        false_A = 15
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            log#_A(x1) = 0
            s_A(x1) = x1 + 10
            quot_A(x1,x2) = x1 + 5
            |0|_A() = 2
            minus_A(x1,x2) = x1
            if_minus_A(x1,x2,x3) = x2
            le_A(x1,x2) = 1
            true_A() = 2
            false_A() = 2
            log#!_A(x1) = x1
            s!_A(x1) = x1 + 16
            quot!_A(x1,x2) = x1 + 11
            |0|!_A() = 19
            minus!_A(x1,x2) = 33
            if_minus!_A(x1,x2,x3) = x2 + x3 + 24
            le!_A(x1,x2) = x1 + x2 + 19
            true!_A() = 20
            false!_A() = 32
  
    partial status:
  
      pi(log#) = []
      pi(s) = []
      pi(quot) = []
      pi(|0|) = []
      pi(minus) = []
      pi(if_minus) = []
      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: 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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = max{1, x1 - 7}
        s_A(x1) = max{58, x1 + 9}
        minus_A(x1,x2) = max{6, x1 + 3}
        le_A(x1,x2) = max{27, x1 + 4}
        |0|_A = 30
        true_A = 33
        false_A = 28
        if_minus_A(x1,x2,x3) = max{14, x1 - 14, x2 + 3}
  
    base order 2:
  
      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) = x1 + 5
            minus_A(x1,x2) = x1 + 1
            le_A(x1,x2) = x2 + 1
            |0|_A() = 2
            true_A() = 1
            false_A() = 1
            if_minus_A(x1,x2,x3) = x2 + 1
            quot#!_A(x1,x2) = 0
            s!_A(x1) = x1 + 1
            minus!_A(x1,x2) = x1 + 8
            le!_A(x1,x2) = x2
            |0|!_A() = 5
            true!_A() = 6
            false!_A() = 3
            if_minus!_A(x1,x2,x3) = x2 + 1
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      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: 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

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{0, x2 - 20}
        false_A = 25
        s_A(x1) = x1 + 19
        minus#_A(x1,x2) = max{0, x1 - 2}
        le_A(x1,x2) = max{9, x1 + 6}
        |0|_A = 39
        true_A = 40
        minus_A(x1,x2) = max{24, x1 + 22, x2 - 41}
  
    base order 2:
  
      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() = 1
            s_A(x1) = x1 + 1
            minus#_A(x1,x2) = x1
            le_A(x1,x2) = x1
            |0|_A() = 3
            true_A() = 1
            minus_A(x1,x2) = x1 + x2 + 2
            if_minus#!_A(x1,x2,x3) = x1 + x3 + 5
            false!_A() = 1
            s!_A(x1) = 5
            minus#!_A(x1,x2) = 0
            le!_A(x1,x2) = x2 + 8
            |0|!_A() = 6
            true!_A() = 7
            minus!_A(x1,x2) = 2
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = []
      pi(minus#) = []
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{10, x2 - 10}
        s_A(x1) = max{21, x1 + 9}
        minus_A(x1,x2) = max{33, x1 + 19, x2 + 20}
        |0|_A = 37
  
    base order 2:
  
      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
            s_A(x1) = x1 + 2
            minus_A(x1,x2) = x1 + x2 + 3
            |0|_A() = 1
            le#!_A(x1,x2) = x1 + x2 + 6
            s!_A(x1) = 5
            minus!_A(x1,x2) = 3
            |0|!_A() = 2
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.