YES
We show the termination of the TRS R:

  minus(minus(x,y),z) -> minus(x,plus(y,z))
  minus(|0|(),y) -> |0|()
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  plus(|0|(),y) -> y
  plus(s(x),y) -> plus(x,s(y))
  plus(s(x),y) -> s(plus(y,x))
  zero(s(x)) -> false()
  zero(|0|()) -> true()
  p(s(x)) -> x
  p(|0|()) -> |0|()
  div(x,y) -> quot(x,y,|0|())
  quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
  quot(|0|(),s(y),z) -> z
  ack(|0|(),x) -> s(x)
  ack(|0|(),x) -> plus(x,s(|0|()))
  ack(s(x),|0|()) -> ack(x,s(|0|()))
  ack(s(x),s(y)) -> ack(x,ack(s(x),y))


-- SCC decomposition.

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

p1: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p2: minus#(minus(x,y),z) -> plus#(y,z)
p3: minus#(s(x),s(y)) -> minus#(x,y)
p4: plus#(s(x),y) -> plus#(x,s(y))
p5: plus#(s(x),y) -> plus#(y,x)
p6: div#(x,y) -> quot#(x,y,|0|())
p7: quot#(s(x),s(y),z) -> quot#(minus(p(ack(|0|(),x)),y),s(y),s(z))
p8: quot#(s(x),s(y),z) -> minus#(p(ack(|0|(),x)),y)
p9: quot#(s(x),s(y),z) -> p#(ack(|0|(),x))
p10: quot#(s(x),s(y),z) -> ack#(|0|(),x)
p11: ack#(|0|(),x) -> plus#(x,s(|0|()))
p12: ack#(s(x),|0|()) -> ack#(x,s(|0|()))
p13: ack#(s(x),s(y)) -> ack#(x,ack(s(x),y))
p14: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The estimated dependency graph contains the following SCCs:

  {p7}
  {p1, p3}
  {p12, p13, p14}
  {p4, p5}


-- Reduction pair.

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

p1: quot#(s(x),s(y),z) -> quot#(minus(p(ack(|0|(),x)),y),s(y),s(z))

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r10, r11, r15, r16

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          quot#_A(x1,x2,x3) = x1 + 1
          s_A(x1) = x1 + 1
          minus_A(x1,x2) = x1
          p_A(x1) = x1
          ack_A(x1,x2) = x2 + 1
          |0|_A() = 0
          plus_A(x1,x2) = x1 + x2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2,x3) = max{1, x1 - 7, x2 - 7, x3 - 1}
          s_A(x1) = x1
          minus_A(x1,x2) = x1
          p_A(x1) = x1
          ack_A(x1,x2) = max{8, x1 - 8, x2}
          |0|_A = 8
          plus_A(x1,x2) = max{1, x1, x2}
          quot#!_A(x1,x2,x3) = max{x1 + 28, x2 + 37, x3 + 37}
          s!_A(x1) = max{16, x1 + 6}
          minus!_A(x1,x2) = max{5, x1 - 8}
          p!_A(x1) = max{4, x1 - 9}
          ack!_A(x1,x2) = max{26, x2 + 7}
          |0|!_A = 2
          plus!_A(x1,x2) = max{x1 + 6, x2 + 17}
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = []
      pi(minus) = [1]
      pi(p) = []
      pi(ack) = []
      pi(|0|) = []
      pi(plus) = [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: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p2: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + 3
          minus_A(x1,x2) = x1 + x2 + 7
          plus_A(x1,x2) = x1 + x2 + 3
          s_A(x1) = x1 + 2
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = max{0, x1 - 42, x2 - 79}
          minus_A(x1,x2) = max{x1 + 41, x2 - 37}
          plus_A(x1,x2) = max{36, x1, x2}
          s_A(x1) = max{3, x1}
          |0|_A = 12
          minus#!_A(x1,x2) = max{3, x1 - 42, x2 - 80}
          minus!_A(x1,x2) = max{2, x1 - 45, x2 - 119}
          plus!_A(x1,x2) = max{1, x1 - 79, x2 - 83}
          s!_A(x1) = max{0, x1 - 83}
          |0|!_A = 0
  
    partial status:
  
      pi(minus#) = [1]
      pi(minus) = [1, 2]
      pi(plus) = [1]
      pi(s) = []
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: ack#(s(x),s(y)) -> ack#(x,ack(s(x),y))
p2: ack#(s(x),s(y)) -> ack#(s(x),y)
p3: ack#(s(x),|0|()) -> ack#(x,s(|0|()))

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r5, r11, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          ack#_A(x1,x2) = 0
          s_A(x1) = 4
          ack_A(x1,x2) = 10
          |0|_A() = 1
          plus_A(x1,x2) = x2 + 5
          p_A(x1) = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ack#_A(x1,x2) = max{0, x1 - 6}
          s_A(x1) = x1 + 5
          ack_A(x1,x2) = max{x1 - 43, x2 + 60}
          |0|_A = 21
          plus_A(x1,x2) = max{61, x1 - 31, x2 + 59}
          p_A(x1) = max{38, x1 - 22}
          ack#!_A(x1,x2) = max{1, x1 - 1}
          s!_A(x1) = max{12, x1 + 4}
          ack!_A(x1,x2) = max{0, x2 - 13}
          |0|!_A = 26
          plus!_A(x1,x2) = max{28, x1 - 27, x2 - 27}
          p!_A(x1) = max{13, x1 - 85}
  
    partial status:
  
      pi(ack#) = []
      pi(s) = []
      pi(ack) = []
      pi(|0|) = []
      pi(plus) = []
      pi(p) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r2, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          ack#_A(x1,x2) = x2 + 1
          s_A(x1) = x1 + 5
          minus_A(x1,x2) = x1 + 1
          |0|_A() = 4
          ack_A(x1,x2) = x1 + x2 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ack#_A(x1,x2) = 0
          s_A(x1) = max{2, x1 - 38}
          minus_A(x1,x2) = max{10, x1 - 12, x2 - 38}
          |0|_A = 3
          ack_A(x1,x2) = max{19, x1 - 25, x2 - 20}
          ack#!_A(x1,x2) = max{4, x1 - 79, x2 - 3}
          s!_A(x1) = max{4, x1 - 118}
          minus!_A(x1,x2) = max{5, x1 - 4}
          |0|!_A = 2
          ack!_A(x1,x2) = max{4, x1 - 80, x2 - 39}
  
    partial status:
  
      pi(ack#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(ack) = [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: plus#(s(x),y) -> plus#(x,s(y))
p2: plus#(s(x),y) -> plus#(y,x)

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 5
          s_A(x1) = x1 + 4
          p_A(x1) = x1 + 1
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{24, x1 - 11}
          p_A(x1) = max{51, x1 - 2}
          |0|_A = 1
          plus#!_A(x1,x2) = max{0, x2 - 123}
          s!_A(x1) = max{5, x1 - 123}
          p!_A(x1) = max{4, x1 - 144}
          |0|!_A = 1
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: minus(minus(x,y),z) -> minus(x,plus(y,z))
r2: minus(|0|(),y) -> |0|()
r3: minus(x,|0|()) -> x
r4: minus(s(x),s(y)) -> minus(x,y)
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> plus(x,s(y))
r7: plus(s(x),y) -> s(plus(y,x))
r8: zero(s(x)) -> false()
r9: zero(|0|()) -> true()
r10: p(s(x)) -> x
r11: p(|0|()) -> |0|()
r12: div(x,y) -> quot(x,y,|0|())
r13: quot(s(x),s(y),z) -> quot(minus(p(ack(|0|(),x)),y),s(y),s(z))
r14: quot(|0|(),s(y),z) -> z
r15: ack(|0|(),x) -> s(x)
r16: ack(|0|(),x) -> plus(x,s(|0|()))
r17: ack(s(x),|0|()) -> ack(x,s(|0|()))
r18: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r11, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 1
          s_A(x1) = x1 + 3
          p_A(x1) = x1 + 1
          |0|_A() = 1
          ack_A(x1,x2) = x1 + x2 + 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x2 - 2}
          s_A(x1) = 1
          p_A(x1) = max{17, x1 + 6}
          |0|_A = 19
          ack_A(x1,x2) = max{11, x1 + 7, x2 - 12}
          plus#!_A(x1,x2) = max{9, x2 + 7}
          s!_A(x1) = max{0, x1 - 40}
          p!_A(x1) = max{10, x1 - 27}
          |0|!_A = 0
          ack!_A(x1,x2) = max{10, x1 - 60, x2}
  
    partial status:
  
      pi(plus#) = [1, 2]
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(ack) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.