YES

We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(|0|(),y) -> |0|()
  minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
  minus(x,plus(y,z)) -> minus(minus(x,y),z)
  p(s(s(x))) -> s(p(s(x)))
  p(|0|()) -> s(s(|0|()))
  div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
  div(plus(x,y),z) -> plus(div(x,z),div(y,z))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(s(x),s(y)) -> minus#(p(s(x)),p(s(y)))
p2: minus#(s(x),s(y)) -> p#(s(x))
p3: minus#(s(x),s(y)) -> p#(s(y))
p4: minus#(x,plus(y,z)) -> minus#(minus(x,y),z)
p5: minus#(x,plus(y,z)) -> minus#(x,y)
p6: p#(s(s(x))) -> p#(s(x))
p7: div#(s(x),s(y)) -> div#(minus(x,y),s(y))
p8: div#(s(x),s(y)) -> minus#(x,y)
p9: div#(plus(x,y),z) -> plus#(div(x,z),div(y,z))
p10: div#(plus(x,y),z) -> div#(x,z)
p11: div#(plus(x,y),z) -> div#(y,z)
p12: plus#(s(x),y) -> plus#(y,minus(s(x),s(|0|())))
p13: plus#(s(x),y) -> minus#(s(x),s(|0|()))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(|0|(),y) -> |0|()
r3: minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
r4: minus(x,plus(y,z)) -> minus(minus(x,y),z)
r5: p(s(s(x))) -> s(p(s(x)))
r6: p(|0|()) -> s(s(|0|()))
r7: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r8: div(plus(x,y),z) -> plus(div(x,z),div(y,z))
r9: plus(|0|(),y) -> y
r10: plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p7, p10, p11}
  {p12}
  {p1, p4, p5}
  {p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(plus(x,y),z) -> div#(y,z)
p2: div#(plus(x,y),z) -> div#(x,z)
p3: div#(s(x),s(y)) -> div#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(|0|(),y) -> |0|()
r3: minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
r4: minus(x,plus(y,z)) -> minus(minus(x,y),z)
r5: p(s(s(x))) -> s(p(s(x)))
r6: p(|0|()) -> s(s(|0|()))
r7: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r8: div(plus(x,y),z) -> plus(div(x,z),div(y,z))
r9: plus(|0|(),y) -> y
r10: plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        div#_A(x1,x2) = ((0,1),(0,1)) x1
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,1)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        minus_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,0)
        p_A(x1) = ((1,0),(1,0)) x1
        |0|_A() = (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        div#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
        s_A(x1) = ((0,1),(0,1)) x1 + (1,4)
        minus_A(x1,x2) = ((1,1),(1,0)) x1 + (1,1)
        p_A(x1) = ((0,0),(1,0)) x1 + (3,0)
        |0|_A() = (1,3)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        div#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
        s_A(x1) = (1,1)
        minus_A(x1,x2) = ((0,0),(1,0)) x1 + (2,1)
        p_A(x1) = (2,1)
        |0|_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(x),y) -> plus#(y,minus(s(x),s(|0|())))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(|0|(),y) -> |0|()
r3: minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
r4: minus(x,plus(y,z)) -> minus(minus(x,y),z)
r5: p(s(s(x))) -> s(p(s(x)))
r6: p(|0|()) -> s(s(|0|()))
r7: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r8: div(plus(x,y),z) -> plus(div(x,z),div(y,z))
r9: plus(|0|(),y) -> y
r10: plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = x1 + x2
        s_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        minus_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        |0|_A() = (1,1)
        plus_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,0)
        p_A(x1) = (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x2
        s_A(x1) = (3,1)
        minus_A(x1,x2) = x1 + (2,0)
        |0|_A() = (1,1)
        plus_A(x1,x2) = x1 + x2 + (1,1)
        p_A(x1) = (4,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,0)) x1
        s_A(x1) = (2,2)
        minus_A(x1,x2) = (0,1)
        |0|_A() = (1,2)
        plus_A(x1,x2) = ((0,1),(0,0)) x2 + (1,1)
        p_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(s(x),s(y)) -> minus#(p(s(x)),p(s(y)))
p2: minus#(x,plus(y,z)) -> minus#(x,y)
p3: minus#(x,plus(y,z)) -> minus#(minus(x,y),z)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(|0|(),y) -> |0|()
r3: minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
r4: minus(x,plus(y,z)) -> minus(minus(x,y),z)
r5: p(s(s(x))) -> s(p(s(x)))
r6: p(|0|()) -> s(s(|0|()))
r7: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r8: div(plus(x,y),z) -> plus(div(x,z),div(y,z))
r9: plus(|0|(),y) -> y
r10: plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = x1 + (1,1)
        p_A(x1) = x1 + (0,1)
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        minus_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (0,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x2
        s_A(x1) = ((1,0),(1,0)) x1 + (2,0)
        p_A(x1) = ((0,1),(0,1)) x1 + (1,0)
        plus_A(x1,x2) = (1,1)
        minus_A(x1,x2) = x1
        |0|_A() = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
        p_A(x1) = (2,2)
        plus_A(x1,x2) = (1,1)
        minus_A(x1,x2) = x1
        |0|_A() = (0,1)
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(|0|(),y) -> |0|()
r3: minus(s(x),s(y)) -> minus(p(s(x)),p(s(y)))
r4: minus(x,plus(y,z)) -> minus(minus(x,y),z)
r5: p(s(s(x))) -> s(p(s(x)))
r6: p(|0|()) -> s(s(|0|()))
r7: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r8: div(plus(x,y),z) -> plus(div(x,z),div(y,z))
r9: plus(|0|(),y) -> y
r10: plus(s(x),y) -> s(plus(y,minus(s(x),s(|0|()))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        p#_A(x1) = x1
        s_A(x1) = x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        p#_A(x1) = x1
        s_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        p#_A(x1) = x1
        s_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.