YES
We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(|0|(),x) -> x
  +(s(x),s(y)) -> s(s(+(x,y)))
  +(+(x,y),z) -> +(x,+(y,z))
  *(x,|0|()) -> |0|()
  *(|0|(),x) -> |0|()
  *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
  *(*(x,y),z) -> *(x,*(y,z))
  app(nil(),l) -> l
  app(cons(x,l1),l2) -> cons(x,app(l1,l2))
  sum(nil()) -> |0|()
  sum(cons(x,l)) -> +(x,sum(l))
  sum(app(l1,l2)) -> +(sum(l1),sum(l2))
  prod(nil()) -> s(|0|())
  prod(cons(x,l)) -> *(x,prod(l))
  prod(app(l1,l2)) -> *(prod(l1),prod(l2))


-- SCC decomposition.

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

p1: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),z) -> +#(y,z)
p4: *#(s(x),s(y)) -> +#(*(x,y),+(x,y))
p5: *#(s(x),s(y)) -> *#(x,y)
p6: *#(s(x),s(y)) -> +#(x,y)
p7: *#(*(x,y),z) -> *#(x,*(y,z))
p8: *#(*(x,y),z) -> *#(y,z)
p9: app#(cons(x,l1),l2) -> app#(l1,l2)
p10: sum#(cons(x,l)) -> +#(x,sum(l))
p11: sum#(cons(x,l)) -> sum#(l)
p12: sum#(app(l1,l2)) -> +#(sum(l1),sum(l2))
p13: sum#(app(l1,l2)) -> sum#(l1)
p14: sum#(app(l1,l2)) -> sum#(l2)
p15: prod#(cons(x,l)) -> *#(x,prod(l))
p16: prod#(cons(x,l)) -> prod#(l)
p17: prod#(app(l1,l2)) -> *#(prod(l1),prod(l2))
p18: prod#(app(l1,l2)) -> prod#(l1)
p19: prod#(app(l1,l2)) -> prod#(l2)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p11, p13, p14}
  {p16, p18, p19}
  {p5, p7, p8}
  {p1, p2, p3}
  {p9}


-- Reduction pair.

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

p1: sum#(app(l1,l2)) -> sum#(l2)
p2: sum#(app(l1,l2)) -> sum#(l1)
p3: sum#(cons(x,l)) -> sum#(l)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

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:
      
        sum#_A(x1) = x1 + 4
        app_A(x1,x2) = max{35, x1 + 6, x2 + 24}
        cons_A(x1,x2) = max{29, x1 - 2, x2 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = 0
          app_A(x1,x2) = max{3, x1 - 25, x2 - 25}
          cons_A(x1,x2) = max{0, x1 - 25, x2 - 27}
          sum#!_A(x1) = 1
          app!_A(x1,x2) = max{7, x1 - 4, x2 - 1}
          cons!_A(x1,x2) = max{6, x1 - 1, x2 - 5}
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [2]
      pi(cons) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: prod#(app(l1,l2)) -> prod#(l2)
p2: prod#(app(l1,l2)) -> prod#(l1)
p3: prod#(cons(x,l)) -> prod#(l)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

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:
      
        prod#_A(x1) = x1 + 4
        app_A(x1,x2) = max{44, x1 + 6, x2 + 8}
        cons_A(x1,x2) = max{38, x1 - 2, x2 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          prod#_A(x1) = 0
          app_A(x1,x2) = max{5, x1 - 70, x2 - 70}
          cons_A(x1,x2) = max{3, x1 - 70, x2 - 71}
          prod#!_A(x1) = 0
          app!_A(x1,x2) = max{5, x1 - 78, x2 - 78}
          cons!_A(x1,x2) = max{4, x1 - 1, x2 - 7}
  
    partial status:
  
      pi(prod#) = [1]
      pi(app) = [2]
      pi(cons) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(*(x,y),z) -> *#(x,*(y,z))
p3: *#(s(x),s(y)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = 1
        *_A(x1,x2) = max{x1 + 6, x2 - 8}
        s_A(x1) = 3
        +_A(x1,x2) = max{x1 + 7, x2}
        |0|_A = 5
        sum_A(x1) = 6
        nil_A = 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = max{0, x1 - 25, x2 - 36}
          *_A(x1,x2) = max{24, x1 + 11, x2}
          s_A(x1) = max{12, x1}
          +_A(x1,x2) = max{24, x1, x2}
          |0|_A = 25
          sum_A(x1) = max{41, x1 - 3}
          nil_A = 0
          *#!_A(x1,x2) = max{11, x1 - 1}
          *!_A(x1,x2) = max{9, x1 - 26, x2 - 51}
          s!_A(x1) = max{1, x1 - 36}
          +!_A(x1,x2) = max{7, x1 - 36, x2 - 36}
          |0|!_A = 2
          sum!_A(x1) = max{6, x1 - 55}
          nil!_A = 3
  
    partial status:
  
      pi(*#) = []
      pi(*) = [1]
      pi(s) = []
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(sum) = []
      pi(nil) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(s(x),s(y)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(s(x),s(y)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{3, x2 + 1}
        *_A(x1,x2) = max{4, x1 - 14, x2 - 15}
        s_A(x1) = x1 + 26
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = max{19, x1 - 20, x2 - 20}
          s_A(x1) = max{6, x1 - 20}
          |0|_A = 17
          *#!_A(x1,x2) = 0
          *!_A(x1,x2) = max{2, x1 - 18}
          s!_A(x1) = max{3, x1 - 84}
          |0|!_A = 1
  
    partial status:
  
      pi(*#) = [2]
      pi(*) = []
      pi(s) = [1]
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

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:
      
        *#_A(x1,x2) = max{35, x1}
        *_A(x1,x2) = max{35, x1 + 17, x2 + 18}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = max{1, x1 - 22, x2 - 25}
          *#!_A(x1,x2) = 0
          *!_A(x1,x2) = max{5, x1 - 3, x2 - 5}
  
    partial status:
  
      pi(*#) = [1]
      pi(*) = [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: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        +#_A(x1,x2) = 0
        s_A(x1) = max{16, x1 - 42}
        +_A(x1,x2) = max{27, x1 + 14, x2 + 14}
        |0|_A = 16
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = max{0, x1 - 14}
          s_A(x1) = max{21, x1 + 10}
          +_A(x1,x2) = max{13, x1 + 10, x2 + 10}
          |0|_A = 21
          +#!_A(x1,x2) = max{0, x1 - 20}
          s!_A(x1) = max{1, x1 - 31}
          +!_A(x1,x2) = max{5, x1 - 48, x2 - 40}
          |0|!_A = 0
  
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(+) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(y,z)
p2: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(y,z)
p2: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

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:
      
        +#_A(x1,x2) = max{25, x1 + 13, x2 - 11}
        +_A(x1,x2) = max{x1 + 12, x2}
        |0|_A = 8
        s_A(x1) = max{14, x1 - 27}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          +#_A(x1,x2) = max{0, x1 - 18, x2 - 26}
          +_A(x1,x2) = max{17, x1 + 8, x2}
          |0|_A = 28
          s_A(x1) = max{28, x1 - 29}
          +#!_A(x1,x2) = max{1, x1 - 18, x2 - 44}
          +!_A(x1,x2) = max{5, x1 - 29, x2 - 37}
          |0|!_A = 0
          s!_A(x1) = max{1, x1 - 66}
  
    partial status:
  
      pi(+#) = [1]
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(s) = []

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: app#(cons(x,l1),l2) -> app#(l1,l2)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(|0|(),x) -> x
r3: +(s(x),s(y)) -> s(s(+(x,y)))
r4: +(+(x,y),z) -> +(x,+(y,z))
r5: *(x,|0|()) -> |0|()
r6: *(|0|(),x) -> |0|()
r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y)))
r8: *(*(x,y),z) -> *(x,*(y,z))
r9: app(nil(),l) -> l
r10: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r11: sum(nil()) -> |0|()
r12: sum(cons(x,l)) -> +(x,sum(l))
r13: sum(app(l1,l2)) -> +(sum(l1),sum(l2))
r14: prod(nil()) -> s(|0|())
r15: prod(cons(x,l)) -> *(x,prod(l))
r16: prod(app(l1,l2)) -> *(prod(l1),prod(l2))

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:
      
        app#_A(x1,x2) = x1 + 12
        cons_A(x1,x2) = max{x1 - 10, x2 + 28}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = 0
          cons_A(x1,x2) = max{153, x1 + 151, x2 - 117}
          app#!_A(x1,x2) = max{x1 + 123, x2 + 6}
          cons!_A(x1,x2) = max{4, x1 - 53, x2 - 55}
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = []

The following rules are removed:

  p1

Then no dependency pair remains.