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, r4, r5, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if#_A(x1,x2,x3,x4,x5) = 0
              true_A() = 1
              le_A(x1,x2) = x1 + 1
              s_A(x1) = x1 + 1
              min_A(x1,x2) = 0
              len_A(x1) = x1 + 1
              cons_A(x1,x2) = x2 + 1
              sum_A(x1,x2) = 1
              take_A(x1,x2) = x1 + 1
              |0|_A() = 0
              nil_A() = 0
              false_A() = 2
              if#!_A(x1,x2,x3,x4,x5) = 5
              true!_A() = 9
              le!_A(x1,x2) = x2 + 9
              s!_A(x1) = 6
              min!_A(x1,x2) = 1
              len!_A(x1) = 7
              cons!_A(x1,x2) = 8
              sum!_A(x1,x2) = 4
              take!_A(x1,x2) = 3
              |0|!_A() = 0
              nil!_A() = 1
              false!_A() = 7
        
        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, x4 - 1}
            true_A = 0
            le_A(x1,x2) = 5
            s_A(x1) = 23
            min_A(x1,x2) = 38
            len_A(x1) = 26
            cons_A(x1,x2) = max{56, x2 + 32}
            sum_A(x1,x2) = max{22, x2 + 2}
            take_A(x1,x2) = x2 + 18
            |0|_A = 24
            nil_A = 11
            false_A = 1
            if#!_A(x1,x2,x3,x4,x5) = max{x2 + 13, x3 + 20, x4 + 20}
            true!_A = 50
            le!_A(x1,x2) = max{49, x1 + 26, x2 - 39}
            s!_A(x1) = x1 + 16
            min!_A(x1,x2) = max{21, x1 - 1}
            len!_A(x1) = 44
            cons!_A(x1,x2) = 43
            sum!_A(x1,x2) = max{0, x2 - 10}
            take!_A(x1,x2) = max{51, x2 - 9}
            |0|!_A = 15
            nil!_A = 16
            false!_A = 13
        
  
    partial status:
  
      pi(if#) = [1]
      pi(true) = []
      pi(le) = []
      pi(s) = [1]
      pi(min) = [2]
      pi(len) = [1]
      pi(cons) = [2]
      pi(sum) = []
      pi(take) = [1]
      pi(|0|) = []
      pi(nil) = []
      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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              min#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              min#!_A(x1,x2) = x1 + x2 + 6
              s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            min#_A(x1,x2) = 0
            s_A(x1) = max{26, x1 - 26}
            min#!_A(x1,x2) = max{1, x2 - 31}
            s!_A(x1) = max{0, x1 - 30}
        
  
    partial status:
  
      pi(min#) = []
      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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              len#_A(x1) = x1
              cons_A(x1,x2) = x2 + 1
              len#!_A(x1) = x1 + 6
              cons!_A(x1,x2) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            len#_A(x1) = 0
            cons_A(x1,x2) = max{27, x1 - 140, x2 - 140}
            len#!_A(x1) = 44
            cons!_A(x1,x2) = max{43, x1 - 1, x2 - 56}
        
  
    partial status:
  
      pi(len#) = [1]
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1,x2) = x2
              s_A(x1) = x1 + 1
              sum#!_A(x1,x2) = x2 + 6
              s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = 0
            s_A(x1) = max{22, x1 + 20}
            sum#!_A(x1,x2) = max{1, x1 - 1}
            s!_A(x1) = max{5, x1 - 4}
        
  
    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

  r8

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              le_A(x1,x2) = 2
              |0|_A() = 7
              true_A() = 1
              le#!_A(x1,x2) = x2 + 3
              s!_A(x1) = 2
              le!_A(x1,x2) = x1 + 3
              |0|!_A() = 1
              true!_A() = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{0, x1 - 11}
            le_A(x1,x2) = max{19, x2 - 42}
            |0|_A = 0
            true_A = 18
            le#!_A(x1,x2) = x2 + 13
            s!_A(x1) = max{17, x1 + 14}
            le!_A(x1,x2) = max{28, x1 - 1, x2 + 10}
            |0|!_A = 29
            true!_A = 28
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(le) = [1]
      pi(|0|) = []
      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: 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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              take#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              cons_A(x1,x2) = x2 + 1
              take#!_A(x1,x2) = 7
              s!_A(x1) = x1 + 6
              cons!_A(x1,x2) = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            take#_A(x1,x2) = x2
            s_A(x1) = max{15, x1 - 45}
            cons_A(x1,x2) = max{8, x1 + 7, x2 + 7}
            take#!_A(x1,x2) = 1
            s!_A(x1) = x1 + 11
            cons!_A(x1,x2) = x1 + 2
        
  
    partial status:
  
      pi(take#) = [2]
      pi(s) = []
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.