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:
  
      max/plus interpretation on natural numbers:
      
        log#_A(x1) = max{11, x1 + 3}
        s_A(x1) = x1 + 4
        quot_A(x1,x2) = x1 + 3
        |0|_A = 12
        pred_A(x1) = max{0, x1 - 3}
        minus_A(x1,x2) = x1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          log#_A(x1) = 0
          s_A(x1) = max{6, x1}
          quot_A(x1,x2) = max{16, x2 - 7}
          |0|_A = 15
          pred_A(x1) = max{18, x1}
          minus_A(x1,x2) = x1 + 18
          log#!_A(x1) = 0
          s!_A(x1) = max{0, x1 - 17}
          quot!_A(x1,x2) = x2 + 7
          |0|!_A = 12
          pred!_A(x1) = 2
          minus!_A(x1,x2) = max{14, x1 + 1, x2 - 5}
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = []
      pi(quot) = []
      pi(|0|) = []
      pi(pred) = []
      pi(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: 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

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, x2 - 10}
        s_A(x1) = x1 + 10
        minus_A(x1,x2) = x1 + 2
        pred_A(x1) = max{1, x1 - 3}
        |0|_A = 13
        quot_A(x1,x2) = max{24, x2 - 46}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 11, x2 - 29}
          s_A(x1) = max{26, x1 + 10}
          minus_A(x1,x2) = max{9, x1 + 3, x2 - 8}
          pred_A(x1) = max{18, x1 - 10}
          |0|_A = 2
          quot_A(x1,x2) = max{25, x1 - 28, x2 - 37}
          quot#!_A(x1,x2) = max{1, x2 - 8}
          s!_A(x1) = max{19, x1 - 28}
          minus!_A(x1,x2) = max{21, x1 - 28, x2 - 27}
          pred!_A(x1) = max{20, x1 - 32}
          |0|!_A = 20
          quot!_A(x1,x2) = max{19, x1 - 64, x2 - 66}
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = []
      pi(minus) = [1]
      pi(pred) = []
      pi(|0|) = []
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = x1 + 5
        s_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = 0
          s_A(x1) = x1 + 7
          minus#!_A(x1,x2) = max{5, x2 - 1}
          s!_A(x1) = max{4, x1 - 35}
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.