YES

We show the termination of the TRS R:

  dbl(|0|()) -> |0|()
  dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
  dbls(nil()) -> nil()
  dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
  sel(|0|(),cons(X,Y)) -> activate(X)
  sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
  indx(nil(),X) -> nil()
  indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
  from(X) -> cons(activate(X),n__from(n__s(activate(X))))
  dbl1(|0|()) -> |01|()
  dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
  sel1(|0|(),cons(X,Y)) -> activate(X)
  sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
  quote(|0|()) -> |01|()
  quote(s(X)) -> s1(quote(activate(X)))
  quote(dbl(X)) -> dbl1(X)
  quote(sel(X,Y)) -> sel1(X,Y)
  s(X) -> n__s(X)
  dbl(X) -> n__dbl(X)
  dbls(X) -> n__dbls(X)
  sel(X1,X2) -> n__sel(X1,X2)
  indx(X1,X2) -> n__indx(X1,X2)
  from(X) -> n__from(X)
  activate(n__s(X)) -> s(X)
  activate(n__dbl(X)) -> dbl(X)
  activate(n__dbls(X)) -> dbls(X)
  activate(n__sel(X1,X2)) -> sel(X1,X2)
  activate(n__indx(X1,X2)) -> indx(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: dbl#(s(X)) -> s#(n__s(n__dbl(activate(X))))
p2: dbl#(s(X)) -> activate#(X)
p3: dbls#(cons(X,Y)) -> activate#(X)
p4: dbls#(cons(X,Y)) -> activate#(Y)
p5: sel#(|0|(),cons(X,Y)) -> activate#(X)
p6: sel#(s(X),cons(Y,Z)) -> sel#(activate(X),activate(Z))
p7: sel#(s(X),cons(Y,Z)) -> activate#(X)
p8: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p9: indx#(cons(X,Y),Z) -> activate#(X)
p10: indx#(cons(X,Y),Z) -> activate#(Z)
p11: indx#(cons(X,Y),Z) -> activate#(Y)
p12: from#(X) -> activate#(X)
p13: dbl1#(s(X)) -> dbl1#(activate(X))
p14: dbl1#(s(X)) -> activate#(X)
p15: sel1#(|0|(),cons(X,Y)) -> activate#(X)
p16: sel1#(s(X),cons(Y,Z)) -> sel1#(activate(X),activate(Z))
p17: sel1#(s(X),cons(Y,Z)) -> activate#(X)
p18: sel1#(s(X),cons(Y,Z)) -> activate#(Z)
p19: quote#(s(X)) -> quote#(activate(X))
p20: quote#(s(X)) -> activate#(X)
p21: quote#(dbl(X)) -> dbl1#(X)
p22: quote#(sel(X,Y)) -> sel1#(X,Y)
p23: activate#(n__s(X)) -> s#(X)
p24: activate#(n__dbl(X)) -> dbl#(X)
p25: activate#(n__dbls(X)) -> dbls#(X)
p26: activate#(n__sel(X1,X2)) -> sel#(X1,X2)
p27: activate#(n__indx(X1,X2)) -> indx#(X1,X2)
p28: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p19}
  {p16}
  {p13}
  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p24, p25, p26, p27, p28}


-- Reduction pair.

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

p1: quote#(s(X)) -> quote#(activate(X))

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quote#_A(x1) = x1
          s_A(x1) = max{3, x1}
          activate_A(x1) = max{2, x1}
          dbl_A(x1) = x1 + 5
          |0|_A = 0
          n__s_A(x1) = max{3, x1}
          n__dbl_A(x1) = x1 + 5
          dbls_A(x1) = x1 + 11
          nil_A = 2
          cons_A(x1,x2) = max{x1 + 3, x2}
          n__dbls_A(x1) = x1 + 11
          sel_A(x1,x2) = max{7, x1 + 6, x2 + 3}
          indx_A(x1,x2) = max{x1 + 8, x2 + 6}
          n__sel_A(x1,x2) = max{7, x1 + 6, x2 + 3}
          n__indx_A(x1,x2) = max{x1 + 8, x2 + 6}
          from_A(x1) = max{9, x1 + 6}
          n__from_A(x1) = max{9, x1 + 6}
    
      precedence:
      
        |0| > dbl = n__dbl > from = n__from > indx = n__indx > dbls = cons = n__dbls > s = n__s > activate = sel > n__sel > quote# > nil
      
      partial status:
    
        pi(quote#) = [1]
        pi(s) = [1]
        pi(activate) = [1]
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = [1]
        pi(dbls) = [1]
        pi(nil) = []
        pi(cons) = []
        pi(n__dbls) = [1]
        pi(sel) = [1]
        pi(indx) = [1, 2]
        pi(n__sel) = [1]
        pi(n__indx) = [1, 2]
        pi(from) = [1]
        pi(n__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quote#_A(x1) = x1
          s_A(x1) = x1
          activate_A(x1) = x1
          dbl_A(x1) = x1 + 11
          |0|_A = 10
          n__s_A(x1) = x1
          n__dbl_A(x1) = 10
          dbls_A(x1) = x1
          nil_A = 8
          cons_A(x1,x2) = 6
          n__dbls_A(x1) = x1
          sel_A(x1,x2) = 10
          indx_A(x1,x2) = max{7, x1 - 10, x2 - 4}
          n__sel_A(x1,x2) = 13
          n__indx_A(x1,x2) = 5
          from_A(x1) = x1 + 6
          n__from_A(x1) = 3
    
      precedence:
      
        indx = from > s = n__s > quote# > dbl > activate > n__dbl = cons > |0| = dbls = nil = n__dbls = sel = n__indx = n__from > n__sel
      
      partial status:
    
        pi(quote#) = [1]
        pi(s) = [1]
        pi(activate) = [1]
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = []
        pi(dbls) = [1]
        pi(nil) = []
        pi(cons) = []
        pi(n__dbls) = [1]
        pi(sel) = []
        pi(indx) = []
        pi(n__sel) = []
        pi(n__indx) = []
        pi(from) = []
        pi(n__from) = []
    

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: sel1#(s(X),cons(Y,Z)) -> sel1#(activate(X),activate(Z))

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel1#_A(x1,x2) = max{x1 + 4, x2 + 7}
          s_A(x1) = max{6, x1}
          cons_A(x1,x2) = max{x1 + 2, x2}
          activate_A(x1) = max{2, x1}
          dbl_A(x1) = x1 + 9
          |0|_A = 8
          n__s_A(x1) = max{6, x1}
          n__dbl_A(x1) = x1 + 9
          dbls_A(x1) = x1 + 13
          nil_A = 1
          n__dbls_A(x1) = x1 + 13
          sel_A(x1,x2) = max{8, x1 + 3, x2 - 1}
          indx_A(x1,x2) = max{10, x1 + 6, x2 + 1}
          n__sel_A(x1,x2) = max{8, x1 + 3, x2 - 1}
          n__indx_A(x1,x2) = max{10, x1 + 6, x2 + 1}
          from_A(x1) = max{14, x1 + 7}
          n__from_A(x1) = max{14, x1 + 7}
    
      precedence:
      
        |0| = nil = from = n__from > dbl = n__dbl > s = n__s = indx = n__indx > sel1# = activate = sel > dbls > cons > n__dbls > n__sel
      
      partial status:
    
        pi(sel1#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = [1]
        pi(dbls) = [1]
        pi(nil) = []
        pi(n__dbls) = [1]
        pi(sel) = []
        pi(indx) = [1]
        pi(n__sel) = []
        pi(n__indx) = [1]
        pi(from) = []
        pi(n__from) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel1#_A(x1,x2) = max{10, x1}
          s_A(x1) = max{52, x1 + 18}
          cons_A(x1,x2) = 36
          activate_A(x1) = max{42, x1 + 18}
          dbl_A(x1) = 42
          |0|_A = 43
          n__s_A(x1) = max{52, x1 + 18}
          n__dbl_A(x1) = 16
          dbls_A(x1) = 65
          nil_A = 45
          n__dbls_A(x1) = x1 + 66
          sel_A(x1,x2) = 24
          indx_A(x1,x2) = 44
          n__sel_A(x1,x2) = 7
          n__indx_A(x1,x2) = 25
          from_A(x1) = 42
          n__from_A(x1) = 18
    
      precedence:
      
        n__dbls = sel = indx = n__sel > n__indx > s = n__s = nil > activate = dbl = |0| = from = n__from > n__dbl > sel1# = cons = dbls
      
      partial status:
    
        pi(sel1#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(dbl) = []
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = []
        pi(dbls) = []
        pi(nil) = []
        pi(n__dbls) = [1]
        pi(sel) = []
        pi(indx) = []
        pi(n__sel) = []
        pi(n__indx) = []
        pi(from) = []
        pi(n__from) = []
    

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: dbl1#(s(X)) -> dbl1#(activate(X))

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          dbl1#_A(x1) = x1
          s_A(x1) = x1
          activate_A(x1) = x1
          dbl_A(x1) = x1 + 2
          |0|_A = 0
          n__s_A(x1) = x1
          n__dbl_A(x1) = x1 + 2
          dbls_A(x1) = max{9, x1 + 6}
          nil_A = 10
          cons_A(x1,x2) = max{x1 + 3, x2}
          n__dbls_A(x1) = max{9, x1 + 6}
          sel_A(x1,x2) = max{1, x2 - 1}
          indx_A(x1,x2) = max{12, x1 - 3, x2 + 11}
          n__sel_A(x1,x2) = max{1, x2 - 1}
          n__indx_A(x1,x2) = max{12, x1 - 3, x2 + 11}
          from_A(x1) = x1 + 3
          n__from_A(x1) = x1 + 3
    
      precedence:
      
        indx = n__indx > dbls = cons = n__dbls = from = n__from > sel = n__sel > dbl = n__dbl > s = n__s > dbl1# = activate = nil > |0|
      
      partial status:
    
        pi(dbl1#) = [1]
        pi(s) = [1]
        pi(activate) = [1]
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = [1]
        pi(dbls) = [1]
        pi(nil) = []
        pi(cons) = []
        pi(n__dbls) = [1]
        pi(sel) = []
        pi(indx) = []
        pi(n__sel) = []
        pi(n__indx) = []
        pi(from) = [1]
        pi(n__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          dbl1#_A(x1) = max{144, x1 + 13}
          s_A(x1) = x1 + 143
          activate_A(x1) = max{133, x1 + 76}
          dbl_A(x1) = 65
          |0|_A = 278
          n__s_A(x1) = x1 + 67
          n__dbl_A(x1) = 64
          dbls_A(x1) = 114
          nil_A = 115
          cons_A(x1,x2) = 110
          n__dbls_A(x1) = 113
          sel_A(x1,x2) = 133
          indx_A(x1,x2) = 110
          n__sel_A(x1,x2) = 56
          n__indx_A(x1,x2) = 109
          from_A(x1) = 110
          n__from_A(x1) = 110
    
      precedence:
      
        n__sel > dbl1# = s = activate = sel > from > indx > dbl > |0| = n__s = n__dbl = dbls = nil = cons = n__dbls = n__indx = n__from
      
      partial status:
    
        pi(dbl1#) = []
        pi(s) = []
        pi(activate) = []
        pi(dbl) = []
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = []
        pi(dbls) = []
        pi(nil) = []
        pi(cons) = []
        pi(n__dbls) = []
        pi(sel) = []
        pi(indx) = []
        pi(n__sel) = []
        pi(n__indx) = []
        pi(from) = []
        pi(n__from) = []
    

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: dbl#(s(X)) -> activate#(X)
p2: activate#(n__from(X)) -> from#(X)
p3: from#(X) -> activate#(X)
p4: activate#(n__indx(X1,X2)) -> indx#(X1,X2)
p5: indx#(cons(X,Y),Z) -> activate#(Y)
p6: activate#(n__sel(X1,X2)) -> sel#(X1,X2)
p7: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p8: activate#(n__dbls(X)) -> dbls#(X)
p9: dbls#(cons(X,Y)) -> activate#(Y)
p10: activate#(n__dbl(X)) -> dbl#(X)
p11: dbls#(cons(X,Y)) -> activate#(X)
p12: sel#(s(X),cons(Y,Z)) -> activate#(X)
p13: sel#(s(X),cons(Y,Z)) -> sel#(activate(X),activate(Z))
p14: sel#(|0|(),cons(X,Y)) -> activate#(X)
p15: indx#(cons(X,Y),Z) -> activate#(Z)
p16: indx#(cons(X,Y),Z) -> activate#(X)

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          dbl#_A(x1) = x1 + 16
          s_A(x1) = max{17, x1}
          activate#_A(x1) = max{15, x1 + 11}
          n__from_A(x1) = max{36, x1 + 18}
          from#_A(x1) = max{15, x1 + 11}
          n__indx_A(x1,x2) = max{19, x1 + 17, x2 + 15}
          indx#_A(x1,x2) = max{15, x1 + 11, x2 + 11}
          cons_A(x1,x2) = max{3, x1, x2}
          n__sel_A(x1,x2) = max{x1 + 10, x2 + 5}
          sel#_A(x1,x2) = max{15, x1 + 11, x2 + 11}
          n__dbls_A(x1) = max{26, x1 + 22}
          dbls#_A(x1) = max{15, x1 + 11}
          n__dbl_A(x1) = x1 + 18
          activate_A(x1) = max{3, x1}
          |0|_A = 5
          dbl_A(x1) = x1 + 18
          n__s_A(x1) = max{17, x1}
          dbls_A(x1) = max{26, x1 + 22}
          nil_A = 0
          sel_A(x1,x2) = max{x1 + 10, x2 + 5}
          indx_A(x1,x2) = max{19, x1 + 17, x2 + 15}
          from_A(x1) = max{36, x1 + 18}
    
      precedence:
      
        nil > sel# = |0| > n__from = dbls# = from > activate# = from# = n__indx = indx# = indx > dbl# = s = n__dbl = dbl = n__s > n__dbls = dbls > cons > activate = sel > n__sel
      
      partial status:
    
        pi(dbl#) = [1]
        pi(s) = []
        pi(activate#) = []
        pi(n__from) = []
        pi(from#) = []
        pi(n__indx) = [1, 2]
        pi(indx#) = [2]
        pi(cons) = []
        pi(n__sel) = [1, 2]
        pi(sel#) = []
        pi(n__dbls) = [1]
        pi(dbls#) = []
        pi(n__dbl) = []
        pi(activate) = [1]
        pi(|0|) = []
        pi(dbl) = []
        pi(n__s) = []
        pi(dbls) = [1]
        pi(nil) = []
        pi(sel) = []
        pi(indx) = [1, 2]
        pi(from) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          dbl#_A(x1) = max{5, x1 + 3}
          s_A(x1) = 17
          activate#_A(x1) = 5
          n__from_A(x1) = 7
          from#_A(x1) = 5
          n__indx_A(x1,x2) = max{x1 + 40, x2 + 22}
          indx#_A(x1,x2) = 5
          cons_A(x1,x2) = 7
          n__sel_A(x1,x2) = max{41, x1 + 4, x2 + 2}
          sel#_A(x1,x2) = 5
          n__dbls_A(x1) = 6
          dbls#_A(x1) = 5
          n__dbl_A(x1) = 16
          activate_A(x1) = max{17, x1}
          |0|_A = 18
          dbl_A(x1) = 17
          n__s_A(x1) = 8
          dbls_A(x1) = 17
          nil_A = 18
          sel_A(x1,x2) = 18
          indx_A(x1,x2) = max{x1 + 40, x2 + 22}
          from_A(x1) = 7
    
      precedence:
      
        from# = n__s = from > n__indx = indx > cons > n__from = activate = |0| = dbl = dbls = nil = sel > s > sel# > dbl# = activate# = indx# = dbls# = n__dbl > n__sel = n__dbls
      
      partial status:
    
        pi(dbl#) = [1]
        pi(s) = []
        pi(activate#) = []
        pi(n__from) = []
        pi(from#) = []
        pi(n__indx) = []
        pi(indx#) = []
        pi(cons) = []
        pi(n__sel) = [2]
        pi(sel#) = []
        pi(n__dbls) = []
        pi(dbls#) = []
        pi(n__dbl) = []
        pi(activate) = [1]
        pi(|0|) = []
        pi(dbl) = []
        pi(n__s) = []
        pi(dbls) = []
        pi(nil) = []
        pi(sel) = []
        pi(indx) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p2, p6, p10, p11

We remove them from the problem.

-- SCC decomposition.

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

p1: dbl#(s(X)) -> activate#(X)
p2: from#(X) -> activate#(X)
p3: activate#(n__indx(X1,X2)) -> indx#(X1,X2)
p4: indx#(cons(X,Y),Z) -> activate#(Y)
p5: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p6: activate#(n__dbls(X)) -> dbls#(X)
p7: dbls#(cons(X,Y)) -> activate#(Y)
p8: sel#(s(X),cons(Y,Z)) -> activate#(X)
p9: sel#(s(X),cons(Y,Z)) -> sel#(activate(X),activate(Z))
p10: sel#(|0|(),cons(X,Y)) -> activate#(X)
p11: indx#(cons(X,Y),Z) -> activate#(Z)
p12: indx#(cons(X,Y),Z) -> activate#(X)

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p9}
  {p3, p4, p6, p7, p11, p12}


-- Reduction pair.

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

p1: sel#(s(X),cons(Y,Z)) -> sel#(activate(X),activate(Z))

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = max{x1 + 8, x2 + 2}
          s_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 6, x2}
          activate_A(x1) = x1
          dbl_A(x1) = x1 + 5
          |0|_A = 5
          n__s_A(x1) = x1
          n__dbl_A(x1) = x1 + 5
          dbls_A(x1) = max{12, x1 + 5}
          nil_A = 1
          n__dbls_A(x1) = max{12, x1 + 5}
          sel_A(x1,x2) = max{x1, x2 + 2}
          indx_A(x1,x2) = max{10, x1 + 9, x2 + 8}
          n__sel_A(x1,x2) = max{x1, x2 + 2}
          n__indx_A(x1,x2) = max{10, x1 + 9, x2 + 8}
          from_A(x1) = x1 + 8
          n__from_A(x1) = x1 + 8
    
      precedence:
      
        dbl = n__dbl > indx = n__indx > dbls = nil = n__dbls > cons = from = n__from > s = n__s > sel# = activate = |0| = sel = n__sel
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = [1]
        pi(n__dbl) = [1]
        pi(dbls) = []
        pi(nil) = []
        pi(n__dbls) = []
        pi(sel) = []
        pi(indx) = [1]
        pi(n__sel) = []
        pi(n__indx) = [1]
        pi(from) = [1]
        pi(n__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = x1
          s_A(x1) = x1 + 34
          cons_A(x1,x2) = 19
          activate_A(x1) = 34
          dbl_A(x1) = max{34, x1}
          |0|_A = 35
          n__s_A(x1) = 0
          n__dbl_A(x1) = 18
          dbls_A(x1) = 24
          nil_A = 25
          n__dbls_A(x1) = 20
          sel_A(x1,x2) = 34
          indx_A(x1,x2) = 19
          n__sel_A(x1,x2) = 17
          n__indx_A(x1,x2) = 19
          from_A(x1) = 5
          n__from_A(x1) = 1
    
      precedence:
      
        sel# = from > s = n__sel = n__from > activate = sel > indx > n__indx > dbl = |0| > dbls = nil = n__dbls > cons > n__dbl > n__s
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = []
        pi(dbl) = [1]
        pi(|0|) = []
        pi(n__s) = []
        pi(n__dbl) = []
        pi(dbls) = []
        pi(nil) = []
        pi(n__dbls) = []
        pi(sel) = []
        pi(indx) = []
        pi(n__sel) = []
        pi(n__indx) = []
        pi(from) = []
        pi(n__from) = []
    

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: activate#(n__indx(X1,X2)) -> indx#(X1,X2)
p2: indx#(cons(X,Y),Z) -> activate#(X)
p3: activate#(n__dbls(X)) -> dbls#(X)
p4: dbls#(cons(X,Y)) -> activate#(Y)
p5: indx#(cons(X,Y),Z) -> activate#(Z)
p6: indx#(cons(X,Y),Z) -> activate#(Y)

and R consists of:

r1: dbl(|0|()) -> |0|()
r2: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r3: dbls(nil()) -> nil()
r4: dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
r5: sel(|0|(),cons(X,Y)) -> activate(X)
r6: sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
r7: indx(nil(),X) -> nil()
r8: indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: dbl1(|0|()) -> |01|()
r11: dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
r12: sel1(|0|(),cons(X,Y)) -> activate(X)
r13: sel1(s(X),cons(Y,Z)) -> sel1(activate(X),activate(Z))
r14: quote(|0|()) -> |01|()
r15: quote(s(X)) -> s1(quote(activate(X)))
r16: quote(dbl(X)) -> dbl1(X)
r17: quote(sel(X,Y)) -> sel1(X,Y)
r18: s(X) -> n__s(X)
r19: dbl(X) -> n__dbl(X)
r20: dbls(X) -> n__dbls(X)
r21: sel(X1,X2) -> n__sel(X1,X2)
r22: indx(X1,X2) -> n__indx(X1,X2)
r23: from(X) -> n__from(X)
r24: activate(n__s(X)) -> s(X)
r25: activate(n__dbl(X)) -> dbl(X)
r26: activate(n__dbls(X)) -> dbls(X)
r27: activate(n__sel(X1,X2)) -> sel(X1,X2)
r28: activate(n__indx(X1,X2)) -> indx(X1,X2)
r29: activate(n__from(X)) -> from(X)
r30: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{4, x1}
          n__indx_A(x1,x2) = max{7, x1, x2 + 5}
          indx#_A(x1,x2) = max{6, x1, x2 + 5}
          cons_A(x1,x2) = max{x1 + 5, x2}
          n__dbls_A(x1) = max{3, x1 + 2}
          dbls#_A(x1) = max{4, x1 + 1}
    
      precedence:
      
        activate# = n__indx = indx# = cons = n__dbls > dbls#
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__indx) = [1, 2]
        pi(indx#) = [1, 2]
        pi(cons) = [1, 2]
        pi(n__dbls) = [1]
        pi(dbls#) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{8, x1 - 3}
          n__indx_A(x1,x2) = max{x1 + 6, x2 + 12}
          indx#_A(x1,x2) = max{x1 - 8, x2 + 8}
          cons_A(x1,x2) = max{x1 + 5, x2 + 7}
          n__dbls_A(x1) = x1 + 6
          dbls#_A(x1) = max{7, x1 + 2}
    
      precedence:
      
        activate# = n__indx = indx# = cons = n__dbls = dbls#
      
      partial status:
    
        pi(activate#) = []
        pi(n__indx) = [2]
        pi(indx#) = [2]
        pi(cons) = [2]
        pi(n__dbls) = [1]
        pi(dbls#) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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