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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if#_A(x1,x2,x3,x4,x5) = x1 + x3 + x4 + 4
          true_A() = 1
          le_A(x1,x2) = x2 + 1
          s_A(x1) = x1 + 2
          min_A(x1,x2) = 0
          len_A(x1) = 3
          cons_A(x1,x2) = 4
          sum_A(x1,x2) = 4
          take_A(x1,x2) = 3
          |0|_A() = 0
          false_A() = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3,x4,x5) = max{20, x1 - 4, x2 - 3, x3 - 2, x4 - 1}
          true_A = 8
          le_A(x1,x2) = 24
          s_A(x1) = 2
          min_A(x1,x2) = 5
          len_A(x1) = max{1, x1 - 20}
          cons_A(x1,x2) = max{19, x2 + 1}
          sum_A(x1,x2) = max{17, x2 - 20}
          take_A(x1,x2) = max{0, x2 - 10}
          |0|_A = 5
          false_A = 0
          if#!_A(x1,x2,x3,x4,x5) = max{x1 - 7, x2 - 3, x3 + 17, x4 + 14}
          true!_A = 29
          le!_A(x1,x2) = max{28, x1 + 25, x2 + 8}
          s!_A(x1) = 0
          min!_A(x1,x2) = 0
          len!_A(x1) = max{0, x1 - 20}
          cons!_A(x1,x2) = 30
          sum!_A(x1,x2) = x2 + 25
          take!_A(x1,x2) = max{31, x1 + 27}
          |0|!_A = 0
          false!_A = 28
  
    partial status:
  
      pi(if#) = [1, 3]
      pi(true) = []
      pi(le) = []
      pi(s) = []
      pi(min) = []
      pi(len) = []
      pi(cons) = []
      pi(sum) = []
      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

  r1, r2, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          min#_A(x1,x2) = x1 + x2 + 11
          s_A(x1) = x1 + 10
          min_A(x1,x2) = 9
          |0|_A() = 8
          le_A(x1,x2) = x1 + 9
          true_A() = 16
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          min#_A(x1,x2) = 0
          s_A(x1) = max{21, x1 - 59}
          min_A(x1,x2) = max{35, x1 - 22, x2 - 24}
          |0|_A = 16
          le_A(x1,x2) = max{77, x1 - 22, x2 - 17}
          true_A = 0
          min#!_A(x1,x2) = 2
          s!_A(x1) = 2
          min!_A(x1,x2) = max{10, x2 - 58}
          |0|!_A = 7
          le!_A(x1,x2) = max{3, x1 - 24, x2 - 101}
          true!_A = 4
  
    partial status:
  
      pi(min#) = []
      pi(s) = [1]
      pi(min) = []
      pi(|0|) = []
      pi(le) = [1]
      pi(true) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          len#_A(x1) = x1
          cons_A(x1,x2) = x1 + x2 + 9
          min_A(x1,x2) = 3
          |0|_A() = 2
          s_A(x1) = x1 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          len#_A(x1) = 0
          cons_A(x1,x2) = max{9, x1 - 2, x2 - 2}
          min_A(x1,x2) = max{8, x1 - 8, x2 - 8}
          |0|_A = 5
          s_A(x1) = max{7, x1 - 18}
          len#!_A(x1) = 1
          cons!_A(x1,x2) = max{52, x1 - 36, x2}
          min!_A(x1,x2) = x1 + 23
          |0|!_A = 3
          s!_A(x1) = 53
  
    partial status:
  
      pi(len#) = [1]
      pi(cons) = []
      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

  r1, r2, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          sum#_A(x1,x2) = x2
          s_A(x1) = x1 + 4
          min_A(x1,x2) = 3
          |0|_A() = 0
          le_A(x1,x2) = x1
          true_A() = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1,x2) = 0
          s_A(x1) = max{7, x1 - 22}
          min_A(x1,x2) = max{2, x1 - 8, x2 - 64}
          |0|_A = 0
          le_A(x1,x2) = max{59, x1 - 8, x2 - 1}
          true_A = 0
          sum#!_A(x1,x2) = 22
          s!_A(x1) = max{50, x1 + 21}
          min!_A(x1,x2) = max{51, x1 + 23, x2 + 28}
          |0|!_A = 49
          le!_A(x1,x2) = max{x1 - 13, x2 + 42}
          true!_A = 48
  
    partial status:
  
      pi(sum#) = []
      pi(s) = [1]
      pi(min) = []
      pi(|0|) = []
      pi(le) = [1]
      pi(true) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = x1 + x2 + 4
          s_A(x1) = x1 + 3
          min_A(x1,x2) = 25
          |0|_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{25, x1 - 77}
          min_A(x1,x2) = max{38, x1 - 32, x2 - 40}
          |0|_A = 31
          le#!_A(x1,x2) = 3
          s!_A(x1) = 2
          min!_A(x1,x2) = max{206, x2 - 166}
          |0|!_A = 5
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = []
      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

  r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          take#_A(x1,x2) = x1 + x2 + 4
          s_A(x1) = x1 + 5
          cons_A(x1,x2) = x1 + x2 + 11
          le_A(x1,x2) = x1 + x2 + 25
          |0|_A() = 0
          true_A() = 24
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          take#_A(x1,x2) = x2
          s_A(x1) = max{9, x1 - 2}
          cons_A(x1,x2) = max{x1 + 21, x2 + 8}
          le_A(x1,x2) = max{66, x1 - 2, x2 + 65}
          |0|_A = 1
          true_A = 0
          take#!_A(x1,x2) = max{6, x2 - 1}
          s!_A(x1) = max{39, x1 - 70}
          cons!_A(x1,x2) = max{41, x1 - 64, x2 - 95}
          le!_A(x1,x2) = max{x1 - 10, x2 + 37}
          |0|!_A = 40
          true!_A = 47
  
    partial status:
  
      pi(take#) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.