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, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        log#_A(x1) = 0
        s_A(x1) = x1 + 4
        quot_A(x1,x2) = x1 + 10
        |0|_A = 0
        pred_A(x1) = max{0, x1 - 1}
        minus_A(x1,x2) = x1
        log_A(x1) = x1 + 1
  
    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) = x1
            s_A(x1) = x1 + 2
            quot_A(x1,x2) = x1 + 1
            |0|_A() = 3
            pred_A(x1) = x1
            minus_A(x1,x2) = x1
            log_A(x1) = x1
            log#!_A(x1) = x1
            s!_A(x1) = 5
            quot!_A(x1,x2) = x2 + 4
            |0|!_A() = 0
            pred!_A(x1) = 3
            minus!_A(x1,x2) = 4
            log!_A(x1) = 12
  
    partial status:
  
      pi(log#) = []
      pi(s) = [1]
      pi(quot) = [1]
      pi(|0|) = []
      pi(pred) = []
      pi(minus) = [1]
      pi(log) = [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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = max{10, x1 + 2}
        s_A(x1) = x1 + 15
        minus_A(x1,x2) = max{12, x1 + 8}
        pred_A(x1) = max{0, x1 - 8}
        |0|_A = 29
  
    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) = x1
            s_A(x1) = x1 + 2
            minus_A(x1,x2) = x1 + 1
            pred_A(x1) = x1
            |0|_A() = 4
            quot#!_A(x1,x2) = 1
            s!_A(x1) = 2
            minus!_A(x1,x2) = 0
            pred!_A(x1) = x1 + 2
            |0|!_A() = 3
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = []
      pi(pred) = []
      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:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{0, x1 - 1}
        s_A(x1) = 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:
            minus#_A(x1,x2) = x2
            s_A(x1) = x1 + 3
            minus#!_A(x1,x2) = x2
            s!_A(x1) = 2
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.