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, >MG)
  
    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 + 4
          quot_A(x1,x2) = x1
          |0|_A() = 2
          pred_A(x1) = x1
          minus_A(x1,x2) = x1
          log_A(x1) = 3
  
    base order 2:
  
      precedence:
      
        s = quot > log# = minus = log > |0| = pred
      
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = []
      pi(quot) = [1]
      pi(|0|) = []
      pi(pred) = []
      pi(minus) = [1]
      pi(log) = []

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, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    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 + 4
          minus_A(x1,x2) = x1 + 1
          pred_A(x1) = x1
          |0|_A() = 1
          quot_A(x1,x2) = 2
          log_A(x1) = 3
  
    base order 2:
  
      precedence:
      
        minus > pred > quot# = s > quot = log > |0|
      
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(pred) = []
      pi(|0|) = []
      pi(quot) = []
      pi(log) = []

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

  r1, r2, r3, r4, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + 6
          s_A(x1) = x1 + 5
          pred_A(x1) = x1
          minus_A(x1,x2) = x1 + 1
          |0|_A() = 1
          quot_A(x1,x2) = x1 + 3
          log_A(x1) = 2
  
    base order 2:
  
      precedence:
      
        minus# > s > quot > minus = log > pred = |0|
      
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(pred) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = [1]
      pi(log) = []

The following rules are removed:

  p1

Then no dependency pair remains.