YES
We show the termination of the TRS R:

  min(|0|(),y) -> |0|()
  min(s(x),|0|()) -> |0|()
  min(s(x),s(y)) -> min(x,y)
  len(nil()) -> |0|()
  len(cons(x,xs)) -> s(len(xs))
  sum(x,|0|()) -> x
  sum(x,s(y)) -> s(sum(x,y))
  le(|0|(),x) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  take(|0|(),cons(y,ys)) -> y
  take(s(x),cons(y,ys)) -> take(x,ys)
  addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
  if(false(),c,x,y,z) -> z
  if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))


-- SCC decomposition.

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

p1: min#(s(x),s(y)) -> min#(x,y)
p2: len#(cons(x,xs)) -> len#(xs)
p3: sum#(x,s(y)) -> sum#(x,y)
p4: le#(s(x),s(y)) -> le#(x,y)
p5: take#(s(x),cons(y,ys)) -> take#(x,ys)
p6: addList#(x,y) -> if#(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
p7: addList#(x,y) -> le#(|0|(),min(len(x),len(y)))
p8: addList#(x,y) -> min#(len(x),len(y))
p9: addList#(x,y) -> len#(x)
p10: addList#(x,y) -> len#(y)
p11: if#(true(),c,xs,ys,z) -> if#(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))
p12: if#(true(),c,xs,ys,z) -> le#(s(c),min(len(xs),len(ys)))
p13: if#(true(),c,xs,ys,z) -> min#(len(xs),len(ys))
p14: if#(true(),c,xs,ys,z) -> len#(xs)
p15: if#(true(),c,xs,ys,z) -> len#(ys)
p16: if#(true(),c,xs,ys,z) -> sum#(take(c,xs),take(c,ys))
p17: if#(true(),c,xs,ys,z) -> take#(c,xs)
p18: if#(true(),c,xs,ys,z) -> take#(c,ys)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The estimated dependency graph contains the following SCCs:

  {p11}
  {p1}
  {p2}
  {p3}
  {p4}
  {p5}


-- Reduction pair.

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

p1: if#(true(),c,xs,ys,z) -> if#(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2, r3, r8, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3,x4,x5) = max{x1 - 128, x2 - 63, x3 + 1, x4 + 2}
            true_A = 71
            le_A(x1,x2) = 130
            s_A(x1) = 64
            min_A(x1,x2) = 67
            len_A(x1) = max{39, x1 - 3}
            cons_A(x1,x2) = x2 + 20
            sum_A(x1,x2) = 17
            take_A(x1,x2) = 20
            |0|_A = 66
            false_A = 1
            if#!_A(x1,x2,x3,x4,x5) = max{x1 - 61, x2 + 5, x3 + 12, x4 + 69}
            true!_A = 15
            le!_A(x1,x2) = 15
            s!_A(x1) = 16
            min!_A(x1,x2) = 3
            len!_A(x1) = max{13, x1 - 3}
            cons!_A(x1,x2) = x1 + 52
            sum!_A(x1,x2) = 18
            take!_A(x1,x2) = 11
            |0|!_A = 2
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3,x4,x5) = max{0, x3 - 1}
            true_A = 131
            le_A(x1,x2) = 131
            s_A(x1) = 125
            min_A(x1,x2) = 139
            len_A(x1) = max{51, x1 + 39}
            cons_A(x1,x2) = max{28, x1 - 40, x2 - 40}
            sum_A(x1,x2) = max{106, x1 + 37, x2 + 39}
            take_A(x1,x2) = max{40, x2 + 35}
            |0|_A = 139
            false_A = 1
            if#!_A(x1,x2,x3,x4,x5) = max{x1 + 25, x3 + 65, x4 + 69}
            true!_A = 129
            le!_A(x1,x2) = max{73, x1 - 9, x2 - 51}
            s!_A(x1) = 2
            min!_A(x1,x2) = 63
            len!_A(x1) = max{66, x1 - 1}
            cons!_A(x1,x2) = max{68, x1 - 105, x2 + 67}
            sum!_A(x1,x2) = 130
            take!_A(x1,x2) = 129
            |0|!_A = 1
            false!_A = 117
        
  
    partial status:
  
      pi(if#) = [1, 4]
      pi(true) = []
      pi(le) = [2]
      pi(s) = [1]
      pi(min) = []
      pi(len) = []
      pi(cons) = [2]
      pi(sum) = [2]
      pi(take) = []
      pi(|0|) = []
      pi(false) = []

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: min#(s(x),s(y)) -> min#(x,y)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            min#_A(x1,x2) = 0
            s_A(x1) = max{19, x1 - 28}
            min#!_A(x1,x2) = max{0, x2 - 157}
            s!_A(x1) = max{11, x1 - 156}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            min#_A(x1,x2) = 0
            s_A(x1) = max{18, x1 - 63}
            min#!_A(x1,x2) = max{2, x1 - 79, x2 - 81}
            s!_A(x1) = max{1, x1 - 80}
        
  
    partial status:
  
      pi(min#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: len#(cons(x,xs)) -> len#(xs)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            len#_A(x1) = x1
            cons_A(x1,x2) = max{x1 + 22, x2 + 1}
            min_A(x1,x2) = max{23, x1 - 18, x2 - 18}
            |0|_A = 17
            s_A(x1) = max{12, x1 - 42}
            len#!_A(x1) = max{1, x1 - 71}
            cons!_A(x1,x2) = max{1, x1 - 67, x2 - 71}
            min!_A(x1,x2) = max{61, x1 - 161, x2 - 161}
            |0|!_A = 58
            s!_A(x1) = max{63, x1 - 42}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            len#_A(x1) = x1
            cons_A(x1,x2) = max{44, x1 + 42, x2 + 1}
            min_A(x1,x2) = max{27, x1 - 25, x2 - 33}
            |0|_A = 24
            s_A(x1) = max{22, x1 - 63}
            len#!_A(x1) = max{1, x1 - 81}
            cons!_A(x1,x2) = max{135, x1 - 2, x2 - 81}
            min!_A(x1,x2) = max{x1 + 69, x2 - 90}
            |0|!_A = 5
            s!_A(x1) = max{1, x1 - 89}
        
  
    partial status:
  
      pi(len#) = [1]
      pi(cons) = [2]
      pi(min) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 33}
            sum#!_A(x1,x2) = 6
            s!_A(x1) = max{5, x1 - 33}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = 0
            s_A(x1) = max{56, x1 - 62}
            sum#!_A(x1,x2) = max{0, x1 - 1}
            s!_A(x1) = max{8, x1 - 83}
        
  
    partial status:
  
      pi(sum#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{107, x1 - 125}
            min_A(x1,x2) = max{3, x1 - 118, x2 - 1}
            |0|_A = 0
            le#!_A(x1,x2) = max{0, x2 - 369}
            s!_A(x1) = max{3, x1 - 368}
            min!_A(x1,x2) = max{5, x1 - 242, x2 - 148}
            |0|!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{73, x1 - 22}
            min_A(x1,x2) = max{77, x1 - 74, x2 - 76}
            |0|_A = 73
            le#!_A(x1,x2) = max{5, x1 - 74, x2 - 74}
            s!_A(x1) = max{4, x1 - 22}
            min!_A(x1,x2) = max{4, x1 - 151, x2 - 156}
            |0|!_A = 4
        
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(min) = []
      pi(|0|) = []

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: take#(s(x),cons(y,ys)) -> take#(x,ys)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            take#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 19}
            cons_A(x1,x2) = max{4, x1 - 1, x2 - 52}
            min_A(x1,x2) = max{11, x1 - 9, x2 - 7}
            |0|_A = 6
            take#!_A(x1,x2) = max{2, x1 - 71}
            s!_A(x1) = max{6, x1 - 70}
            cons!_A(x1,x2) = max{1, x1 - 12, x2 - 115}
            min!_A(x1,x2) = max{5, x1 - 21, x2 - 23}
            |0|!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            take#_A(x1,x2) = x2
            s_A(x1) = max{22, x1 - 69}
            cons_A(x1,x2) = max{x1 + 1, x2 + 23}
            min_A(x1,x2) = max{38, x1 - 25, x2 - 27}
            |0|_A = 24
            take#!_A(x1,x2) = max{8, x1 - 23, x2 - 46}
            s!_A(x1) = max{2, x1 - 69}
            cons!_A(x1,x2) = max{7, x1 - 68, x2 - 47}
            min!_A(x1,x2) = max{5, x1 - 110, x2 - 105}
            |0|!_A = 2
        
  
    partial status:
  
      pi(take#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(min) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.