YES

We show the termination of the TRS R:

  a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
  a__sqr(|0|()) -> |0|()
  a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
  a__dbl(|0|()) -> |0|()
  a__dbl(s(X)) -> s(s(dbl(X)))
  a__add(|0|(),X) -> mark(X)
  a__add(s(X),Y) -> s(add(X,Y))
  a__first(|0|(),X) -> nil()
  a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
  mark(terms(X)) -> a__terms(mark(X))
  mark(sqr(X)) -> a__sqr(mark(X))
  mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
  mark(dbl(X)) -> a__dbl(mark(X))
  mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(recip(X)) -> recip(mark(X))
  mark(s(X)) -> s(X)
  mark(|0|()) -> |0|()
  mark(nil()) -> nil()
  a__terms(X) -> terms(X)
  a__sqr(X) -> sqr(X)
  a__add(X1,X2) -> add(X1,X2)
  a__dbl(X) -> dbl(X)
  a__first(X1,X2) -> first(X1,X2)

-- SCC decomposition.

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

p1: a__terms#(N) -> a__sqr#(mark(N))
p2: a__terms#(N) -> mark#(N)
p3: a__add#(|0|(),X) -> mark#(X)
p4: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p5: mark#(terms(X)) -> a__terms#(mark(X))
p6: mark#(terms(X)) -> mark#(X)
p7: mark#(sqr(X)) -> a__sqr#(mark(X))
p8: mark#(sqr(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> mark#(X2)
p12: mark#(dbl(X)) -> a__dbl#(mark(X))
p13: mark#(dbl(X)) -> mark#(X)
p14: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p15: mark#(first(X1,X2)) -> mark#(X1)
p16: mark#(first(X1,X2)) -> mark#(X2)
p17: mark#(cons(X1,X2)) -> mark#(X1)
p18: mark#(recip(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p8, p9, p10, p11, p13, p14, p15, p16, p17, p18}


-- Reduction pair.

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

p1: a__terms#(N) -> mark#(N)
p2: mark#(recip(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(first(X1,X2)) -> mark#(X1)
p6: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p7: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> mark#(X2)
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p12: a__add#(|0|(),X) -> mark#(X)
p13: mark#(sqr(X)) -> mark#(X)
p14: mark#(terms(X)) -> mark#(X)
p15: mark#(terms(X)) -> a__terms#(mark(X))

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__terms#_A(x1) = x1 + 42
        mark#_A(x1) = max{42, x1 + 21}
        recip_A(x1) = max{189, x1 + 22}
        cons_A(x1,x2) = max{42, x1 + 32}
        first_A(x1,x2) = max{39, x1 + 30, x2 + 23}
        a__first#_A(x1,x2) = max{31, x1 - 10, x2}
        mark_A(x1) = max{31, x1 + 22}
        s_A(x1) = 52
        dbl_A(x1) = max{78, x1 + 51}
        add_A(x1,x2) = max{94, x1 + 85, x2 + 75}
        a__add#_A(x1,x2) = max{x1 + 10, x2 + 43}
        |0|_A = 33
        sqr_A(x1) = max{111, x1 + 102}
        terms_A(x1) = max{220, x1 + 188}
        a__terms_A(x1) = max{222, x1 + 188}
        a__sqr_A(x1) = max{127, x1 + 102}
        a__dbl_A(x1) = max{100, x1 + 51}
        a__add_A(x1,x2) = max{96, x1 + 85, x2 + 75}
        a__first_A(x1,x2) = max{41, x1 + 30, x2 + 23}
        nil_A = 42
  
    precedence:
    
      a__terms# > mark# = recip = cons = first = a__first# = mark = s = dbl = add = a__add# = |0| = sqr = terms = a__terms = a__sqr = a__dbl = a__add = a__first = nil
    
    partial status:
  
      pi(a__terms#) = []
      pi(mark#) = [1]
      pi(recip) = [1]
      pi(cons) = [1]
      pi(first) = [2]
      pi(a__first#) = [2]
      pi(mark) = [1]
      pi(s) = []
      pi(dbl) = [1]
      pi(add) = []
      pi(a__add#) = [2]
      pi(|0|) = []
      pi(sqr) = [1]
      pi(terms) = []
      pi(a__terms) = []
      pi(a__sqr) = [1]
      pi(a__dbl) = [1]
      pi(a__add) = []
      pi(a__first) = [2]
      pi(nil) = []

The next rules are strictly ordered:

  p15

We remove them from the problem.

-- SCC decomposition.

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

p1: a__terms#(N) -> mark#(N)
p2: mark#(recip(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(first(X1,X2)) -> mark#(X1)
p6: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p7: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> mark#(X2)
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p12: a__add#(|0|(),X) -> mark#(X)
p13: mark#(sqr(X)) -> mark#(X)
p14: mark#(terms(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14}


-- Reduction pair.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(terms(X)) -> mark#(X)
p3: mark#(sqr(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)
p7: mark#(add(X1,X2)) -> mark#(X2)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p10: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p11: mark#(first(X1,X2)) -> mark#(X1)
p12: mark#(first(X1,X2)) -> mark#(X2)
p13: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{33, x1}
        recip_A(x1) = x1 + 8
        terms_A(x1) = max{34, x1 + 33}
        sqr_A(x1) = x1 + 9
        add_A(x1,x2) = max{34, x1, x2 + 8}
        a__add#_A(x1,x2) = max{x1 - 7, x2 + 1}
        mark_A(x1) = x1 + 7
        |0|_A = 41
        dbl_A(x1) = x1 + 1
        first_A(x1,x2) = max{x1 + 85, x2 + 85}
        a__first#_A(x1,x2) = max{85, x1 + 78, x2 + 77}
        s_A(x1) = max{42, x1}
        cons_A(x1,x2) = max{x1 + 8, x2 - 43}
        a__terms_A(x1) = max{39, x1 + 33}
        a__sqr_A(x1) = max{10, x1 + 9}
        a__dbl_A(x1) = max{8, x1 + 1}
        a__add_A(x1,x2) = max{41, x1, x2 + 8}
        a__first_A(x1,x2) = max{92, x1 + 85, x2 + 85}
        nil_A = 84
  
    precedence:
    
      mark > mark# = recip = terms = sqr = add = a__add# = |0| = dbl = first = s = cons = a__terms = a__sqr = a__dbl = a__add = a__first > a__first# = nil
    
    partial status:
  
      pi(mark#) = [1]
      pi(recip) = [1]
      pi(terms) = [1]
      pi(sqr) = [1]
      pi(add) = [1]
      pi(a__add#) = [2]
      pi(mark) = [1]
      pi(|0|) = []
      pi(dbl) = []
      pi(first) = [2]
      pi(a__first#) = []
      pi(s) = []
      pi(cons) = [1]
      pi(a__terms) = [1]
      pi(a__sqr) = [1]
      pi(a__dbl) = []
      pi(a__add) = [1]
      pi(a__first) = [2]
      pi(nil) = []

The next rules are strictly ordered:

  p12

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(terms(X)) -> mark#(X)
p3: mark#(sqr(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)
p7: mark#(add(X1,X2)) -> mark#(X2)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p10: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p11: mark#(first(X1,X2)) -> mark#(X1)
p12: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p5: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p6: mark#(dbl(X)) -> mark#(X)
p7: mark#(add(X1,X2)) -> mark#(X2)
p8: mark#(add(X1,X2)) -> mark#(X1)
p9: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p10: a__add#(|0|(),X) -> mark#(X)
p11: mark#(sqr(X)) -> mark#(X)
p12: mark#(terms(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{32, x1 - 4}
        recip_A(x1) = max{1, x1}
        cons_A(x1,x2) = max{33, x1 + 3}
        first_A(x1,x2) = max{x1 + 109, x2 + 97}
        a__first#_A(x1,x2) = max{92, x1 + 42, x2 + 30}
        mark_A(x1) = x1 + 62
        s_A(x1) = 19
        dbl_A(x1) = max{19, x1 + 4}
        add_A(x1,x2) = max{x1 + 70, x2 + 100}
        a__add#_A(x1,x2) = max{65, x1 + 1, x2 + 33}
        |0|_A = 32
        sqr_A(x1) = max{58, x1}
        terms_A(x1) = x1 + 124
        a__terms_A(x1) = x1 + 124
        a__sqr_A(x1) = max{120, x1}
        a__dbl_A(x1) = max{19, x1 + 4}
        a__add_A(x1,x2) = max{x1 + 70, x2 + 100}
        a__first_A(x1,x2) = max{171, x1 + 109, x2 + 97}
        nil_A = 171
  
    precedence:
    
      cons = mark = a__first > recip = first = nil > mark# = s = dbl = a__add# = sqr = a__terms = a__sqr = a__dbl > a__first# > |0| = terms > add = a__add
    
    partial status:
  
      pi(mark#) = []
      pi(recip) = []
      pi(cons) = []
      pi(first) = []
      pi(a__first#) = [2]
      pi(mark) = []
      pi(s) = []
      pi(dbl) = []
      pi(add) = [1]
      pi(a__add#) = []
      pi(|0|) = []
      pi(sqr) = []
      pi(terms) = []
      pi(a__terms) = []
      pi(a__sqr) = []
      pi(a__dbl) = []
      pi(a__add) = [1]
      pi(a__first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p5: mark#(dbl(X)) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X2)
p7: mark#(add(X1,X2)) -> mark#(X1)
p8: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p9: a__add#(|0|(),X) -> mark#(X)
p10: mark#(sqr(X)) -> mark#(X)
p11: mark#(terms(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(terms(X)) -> mark#(X)
p3: mark#(sqr(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)
p7: mark#(add(X1,X2)) -> mark#(X2)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(first(X1,X2)) -> mark#(X1)
p10: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{3, x1 - 5}
        recip_A(x1) = x1 + 4
        terms_A(x1) = x1 + 106
        sqr_A(x1) = x1 + 62
        add_A(x1,x2) = max{x1, x2 + 31}
        a__add#_A(x1,x2) = max{0, x1 - 36, x2 - 5}
        mark_A(x1) = x1 + 31
        |0|_A = 40
        dbl_A(x1) = x1 + 31
        first_A(x1,x2) = max{x1 + 32, x2 + 42}
        cons_A(x1,x2) = x1 + 9
        a__terms_A(x1) = x1 + 106
        a__sqr_A(x1) = x1 + 62
        s_A(x1) = max{31, x1}
        a__dbl_A(x1) = max{32, x1 + 31}
        a__add_A(x1,x2) = max{x1, x2 + 31}
        a__first_A(x1,x2) = max{43, x1 + 32, x2 + 42}
        nil_A = 41
  
    precedence:
    
      mark# = recip = terms = sqr = add = a__add# = mark = |0| = dbl = first = cons = a__terms = a__sqr = s = a__dbl = a__add = a__first = nil
    
    partial status:
  
      pi(mark#) = []
      pi(recip) = []
      pi(terms) = []
      pi(sqr) = []
      pi(add) = [2]
      pi(a__add#) = []
      pi(mark) = [1]
      pi(|0|) = []
      pi(dbl) = [1]
      pi(first) = [1, 2]
      pi(cons) = []
      pi(a__terms) = []
      pi(a__sqr) = []
      pi(s) = []
      pi(a__dbl) = [1]
      pi(a__add) = [2]
      pi(a__first) = [1, 2]
      pi(nil) = []

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(terms(X)) -> mark#(X)
p3: mark#(sqr(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)
p7: mark#(add(X1,X2)) -> mark#(X2)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(first(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(first(X1,X2)) -> mark#(X1)
p3: mark#(dbl(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p7: a__add#(|0|(),X) -> mark#(X)
p8: mark#(sqr(X)) -> mark#(X)
p9: mark#(terms(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{0, x1 - 6}
        recip_A(x1) = max{18, x1}
        first_A(x1,x2) = max{x1, x2 + 27}
        dbl_A(x1) = max{51, x1}
        add_A(x1,x2) = max{x1 + 31, x2 + 33}
        a__add#_A(x1,x2) = max{x1 - 1, x2 + 1}
        mark_A(x1) = x1 + 25
        |0|_A = 36
        sqr_A(x1) = x1 + 34
        terms_A(x1) = x1 + 74
        a__terms_A(x1) = x1 + 74
        cons_A(x1,x2) = x1 + 14
        a__sqr_A(x1) = max{35, x1 + 34}
        s_A(x1) = 16
        a__dbl_A(x1) = max{76, x1}
        a__add_A(x1,x2) = max{39, x1 + 31, x2 + 33}
        a__first_A(x1,x2) = max{28, x1, x2 + 27}
        nil_A = 23
  
    precedence:
    
      mark# = a__add# > recip = add = mark = |0| = a__terms = s = a__dbl = a__add > dbl = terms > first = sqr = cons = a__sqr = a__first = nil
    
    partial status:
  
      pi(mark#) = []
      pi(recip) = []
      pi(first) = []
      pi(dbl) = []
      pi(add) = []
      pi(a__add#) = []
      pi(mark) = [1]
      pi(|0|) = []
      pi(sqr) = [1]
      pi(terms) = []
      pi(a__terms) = []
      pi(cons) = []
      pi(a__sqr) = [1]
      pi(s) = []
      pi(a__dbl) = []
      pi(a__add) = []
      pi(a__first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p9

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(first(X1,X2)) -> mark#(X1)
p3: mark#(dbl(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p7: a__add#(|0|(),X) -> mark#(X)
p8: mark#(sqr(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: mark#(recip(X)) -> mark#(X)
p2: mark#(sqr(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p4: a__add#(|0|(),X) -> mark#(X)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> mark#(X2)
p7: mark#(dbl(X)) -> mark#(X)
p8: mark#(first(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{7, x1}
        recip_A(x1) = x1 + 19
        sqr_A(x1) = x1 + 18
        add_A(x1,x2) = max{19, x1 + 7, x2 + 9}
        a__add#_A(x1,x2) = max{19, x1 + 6, x2 + 6}
        mark_A(x1) = x1
        |0|_A = 19
        dbl_A(x1) = x1 + 23
        first_A(x1,x2) = max{42, x1, x2 + 13}
        a__terms_A(x1) = x1 + 40
        cons_A(x1,x2) = max{16, x1 + 2}
        a__sqr_A(x1) = x1 + 18
        terms_A(x1) = x1 + 40
        s_A(x1) = 0
        a__dbl_A(x1) = x1 + 23
        a__add_A(x1,x2) = max{19, x1 + 7, x2 + 9}
        a__first_A(x1,x2) = max{42, x1, x2 + 13}
        nil_A = 20
  
    precedence:
    
      sqr = |0| = dbl = a__terms = cons = a__sqr = terms = s = a__dbl = nil > recip = mark = a__add > a__first > first > add > mark# = a__add#
    
    partial status:
  
      pi(mark#) = [1]
      pi(recip) = []
      pi(sqr) = []
      pi(add) = [1]
      pi(a__add#) = []
      pi(mark) = [1]
      pi(|0|) = []
      pi(dbl) = []
      pi(first) = [1]
      pi(a__terms) = []
      pi(cons) = []
      pi(a__sqr) = []
      pi(terms) = []
      pi(s) = []
      pi(a__dbl) = []
      pi(a__add) = [1]
      pi(a__first) = [1]
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(sqr(X)) -> mark#(X)
p2: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p3: a__add#(|0|(),X) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> mark#(X2)
p6: mark#(dbl(X)) -> mark#(X)
p7: mark#(first(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: mark#(sqr(X)) -> mark#(X)
p2: mark#(first(X1,X2)) -> mark#(X1)
p3: mark#(dbl(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p7: a__add#(|0|(),X) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1
        sqr_A(x1) = x1 + 29
        first_A(x1,x2) = max{x1, x2 + 11}
        dbl_A(x1) = max{4, x1}
        add_A(x1,x2) = max{x1 + 7, x2 + 12}
        a__add#_A(x1,x2) = max{x1 - 5, x2}
        mark_A(x1) = x1
        |0|_A = 5
        a__terms_A(x1) = max{49, x1 + 44}
        cons_A(x1,x2) = x1 + 13
        recip_A(x1) = x1 + 2
        a__sqr_A(x1) = x1 + 29
        terms_A(x1) = max{49, x1 + 44}
        s_A(x1) = 4
        a__dbl_A(x1) = max{4, x1}
        a__add_A(x1,x2) = max{x1 + 7, x2 + 12}
        a__first_A(x1,x2) = max{x1, x2 + 11}
        nil_A = 0
  
    precedence:
    
      mark# = sqr = first = dbl = add = a__add# = mark = |0| = a__terms = cons = recip = a__sqr = terms = s = a__dbl = a__add = a__first = nil
    
    partial status:
  
      pi(mark#) = []
      pi(sqr) = [1]
      pi(first) = []
      pi(dbl) = [1]
      pi(add) = []
      pi(a__add#) = [2]
      pi(mark) = [1]
      pi(|0|) = []
      pi(a__terms) = [1]
      pi(cons) = []
      pi(recip) = [1]
      pi(a__sqr) = [1]
      pi(terms) = [1]
      pi(s) = []
      pi(a__dbl) = [1]
      pi(a__add) = []
      pi(a__first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(dbl(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p6: a__add#(|0|(),X) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p3: a__add#(|0|(),X) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> mark#(X2)
p6: mark#(dbl(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{11, x1}
        first_A(x1,x2) = max{x1 + 12, x2 + 15}
        add_A(x1,x2) = max{x1 + 27, x2 + 25}
        a__add#_A(x1,x2) = max{x1 + 1, x2 + 11}
        mark_A(x1) = x1 + 14
        |0|_A = 10
        dbl_A(x1) = x1 + 15
        a__terms_A(x1) = max{155, x1 + 91}
        cons_A(x1,x2) = 71
        recip_A(x1) = x1 + 15
        a__sqr_A(x1) = max{62, x1 + 61}
        terms_A(x1) = max{141, x1 + 91}
        s_A(x1) = 5
        sqr_A(x1) = x1 + 61
        a__dbl_A(x1) = max{16, x1 + 15}
        a__add_A(x1,x2) = max{x1 + 27, x2 + 25}
        a__first_A(x1,x2) = max{x1 + 12, x2 + 15}
        nil_A = 9
  
    precedence:
    
      add = mark = |0| = dbl = a__terms = recip = a__sqr = terms = s = sqr = a__dbl = a__add > a__add# > mark# = first = cons = a__first = nil
    
    partial status:
  
      pi(mark#) = [1]
      pi(first) = []
      pi(add) = [1, 2]
      pi(a__add#) = [1, 2]
      pi(mark) = [1]
      pi(|0|) = []
      pi(dbl) = [1]
      pi(a__terms) = [1]
      pi(cons) = []
      pi(recip) = []
      pi(a__sqr) = [1]
      pi(terms) = [1]
      pi(s) = []
      pi(sqr) = [1]
      pi(a__dbl) = [1]
      pi(a__add) = [1, 2]
      pi(a__first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(dbl(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4, p5}


-- Reduction pair.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(dbl(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        first_A(x1,x2) = max{x1 + 1, x2 + 1}
        dbl_A(x1) = x1
        add_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = first = dbl = add
    
    partial status:
  
      pi(mark#) = []
      pi(first) = [2]
      pi(dbl) = [1]
      pi(add) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(dbl(X)) -> mark#(X)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: mark#(dbl(X)) -> mark#(X)
p2: mark#(add(X1,X2)) -> mark#(X1)
p3: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        dbl_A(x1) = x1 + 1
        add_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      mark# = dbl = add
    
    partial status:
  
      pi(mark#) = []
      pi(dbl) = [1]
      pi(add) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(add(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: mark#(add(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        add_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      mark# = add
    
    partial status:
  
      pi(mark#) = []
      pi(add) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        add_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      mark# = add
    
    partial status:
  
      pi(mark#) = []
      pi(add) = [2]

The next rules are strictly ordered:

  p1

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