YES
We show the termination of the TRS R:

  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(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: minus#(x,s(y)) -> pred#(minus(x,y))
p2: minus#(x,s(y)) -> minus#(x,y)
p3: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p4: quot#(s(x),s(y)) -> minus#(x,y)
p5: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|())))))
p6: log#(s(s(x))) -> quot#(x,s(s(|0|())))

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p3}
  {p2}


-- 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: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2, r3, r4, r5

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!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = 0
            s_A(x1) = max{1, x1}
            quot_A(x1,x2) = 5
            |0|_A = 1
            pred_A(x1) = max{3, x1}
            minus_A(x1,x2) = x1 + 3
            log#!_A(x1) = max{14, x1 + 9}
            s!_A(x1) = max{0, x1 - 9}
            quot!_A(x1,x2) = 0
            |0|!_A = 0
            pred!_A(x1) = 0
            minus!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = 0
            s_A(x1) = max{30, x1 + 17}
            quot_A(x1,x2) = max{5, x1 + 3}
            |0|_A = 29
            pred_A(x1) = x1
            minus_A(x1,x2) = x1
            log#!_A(x1) = max{8, x1 - 21}
            s!_A(x1) = 7
            quot!_A(x1,x2) = max{1, x1 - 13}
            |0|!_A = 7
            pred!_A(x1) = 17
            minus!_A(x1,x2) = max{x1 + 5, x2 - 1}
        
  
    partial status:
  
      pi(log#) = []
      pi(s) = [1]
      pi(quot) = []
      pi(|0|) = []
      pi(pred) = [1]
      pi(minus) = [1, 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: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2, r3

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!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 11, x2 - 17}
            s_A(x1) = max{14, x1 + 10}
            minus_A(x1,x2) = max{9, x1 + 6}
            pred_A(x1) = max{7, x1 - 10}
            |0|_A = 0
            quot#!_A(x1,x2) = x1 + 10
            s!_A(x1) = 0
            minus!_A(x1,x2) = max{23, x1 + 21, x2 - 15}
            pred!_A(x1) = max{22, x1 + 10}
            |0|!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 15, x2 - 29}
            s_A(x1) = max{26, x1 + 14}
            minus_A(x1,x2) = max{13, x1 + 11, x2 - 28}
            pred_A(x1) = max{12, x1 - 14}
            |0|_A = 25
            quot#!_A(x1,x2) = max{2, x2 - 11}
            s!_A(x1) = max{1, x1 - 21}
            minus!_A(x1,x2) = max{16, x1 - 2, x2 + 4}
            pred!_A(x1) = max{17, x1 - 14}
            |0|!_A = 3
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1, 2]
      pi(pred) = [1]
      pi(|0|) = []

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: minus#(x,s(y)) -> minus#(x,y)

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r6: log(s(|0|())) -> |0|()
r7: 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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 4}
            minus#!_A(x1,x2) = 0
            s!_A(x1) = max{2, x1 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 4}
            minus#!_A(x1,x2) = 0
            s!_A(x1) = max{2, x1 - 52}
        
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.