YES
We show the termination of the TRS R:

  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(app(sumwith(),f),nil()) -> nil()
  app(app(sumwith(),f),app(app(cons(),x),xs)) -> app(app(plus(),app(f,x)),app(app(sumwith(),f),xs))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(sumwith_0(),y) -> sumwith_1(y)
  app(sumwith_1(x1),y) -> sumwith_2(x1,y)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  sumwith_2(f,nil_0()) -> nil_0()
  sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))


-- SCC decomposition.

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

p1: app#(plus_1(x1),y) -> plus_2#(x1,y)
p2: app#(sumwith_1(x1),y) -> sumwith_2#(x1,y)
p3: plus_2#(s_1(x),y) -> plus_2#(x,y)
p4: sumwith_2#(f,cons_2(x,xs)) -> plus_2#(app(f,x),sumwith_2(f,xs))
p5: sumwith_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: sumwith_2#(f,cons_2(x,xs)) -> sumwith_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(plus_0(),y) -> plus_1(y)
r4: app(plus_1(x1),y) -> plus_2(x1,y)
r5: app(s_0(),y) -> s_1(y)
r6: app(sumwith_0(),y) -> sumwith_1(y)
r7: app(sumwith_1(x1),y) -> sumwith_2(x1,y)
r8: plus_2(|0|_0(),y) -> y
r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r10: sumwith_2(f,nil_0()) -> nil_0()
r11: sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p6}
  {p3}


-- Reduction pair.

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

p1: sumwith_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(sumwith_1(x1),y) -> sumwith_2#(x1,y)
p3: sumwith_2#(f,cons_2(x,xs)) -> sumwith_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(plus_0(),y) -> plus_1(y)
r4: app(plus_1(x1),y) -> plus_2(x1,y)
r5: app(s_0(),y) -> s_1(y)
r6: app(sumwith_0(),y) -> sumwith_1(y)
r7: app(sumwith_1(x1),y) -> sumwith_2(x1,y)
r8: plus_2(|0|_0(),y) -> y
r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r10: sumwith_2(f,nil_0()) -> nil_0()
r11: sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))

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:
          
            sumwith_2#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            cons_2_A(x1,x2) = max{2, x1 + 1, x2}
            app#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            sumwith_1_A(x1) = x1 + 2
            sumwith_2#!_A(x1,x2) = max{3, x1 - 17, x2 - 3}
            cons_2!_A(x1,x2) = max{2, x2 - 14}
            app#!_A(x1,x2) = max{3, x1 - 18, x2 - 3}
            sumwith_1!_A(x1) = max{1, x1 - 17}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sumwith_2#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
            cons_2_A(x1,x2) = max{1, x1, x2}
            app#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
            sumwith_1_A(x1) = max{1, x1}
            sumwith_2#!_A(x1,x2) = max{3, x1 - 2, x2 - 2}
            cons_2!_A(x1,x2) = max{2, x1 - 9, x2 - 16}
            app#!_A(x1,x2) = max{3, x1 - 2, x2 - 2}
            sumwith_1!_A(x1) = 4
        
  
    partial status:
  
      pi(sumwith_2#) = [1]
      pi(cons_2) = [2]
      pi(app#) = [1]
      pi(sumwith_1) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: sumwith_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: sumwith_2#(f,cons_2(x,xs)) -> sumwith_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(plus_0(),y) -> plus_1(y)
r4: app(plus_1(x1),y) -> plus_2(x1,y)
r5: app(s_0(),y) -> s_1(y)
r6: app(sumwith_0(),y) -> sumwith_1(y)
r7: app(sumwith_1(x1),y) -> sumwith_2(x1,y)
r8: plus_2(|0|_0(),y) -> y
r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r10: sumwith_2(f,nil_0()) -> nil_0()
r11: sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: sumwith_2#(f,cons_2(x,xs)) -> sumwith_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(plus_0(),y) -> plus_1(y)
r4: app(plus_1(x1),y) -> plus_2(x1,y)
r5: app(s_0(),y) -> s_1(y)
r6: app(sumwith_0(),y) -> sumwith_1(y)
r7: app(sumwith_1(x1),y) -> sumwith_2(x1,y)
r8: plus_2(|0|_0(),y) -> y
r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r10: sumwith_2(f,nil_0()) -> nil_0()
r11: sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))

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:
          
            sumwith_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = max{x1 - 7, x2 + 1}
            sumwith_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 2}
            cons_2!_A(x1,x2) = max{2, x1 - 12, x2 - 11}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sumwith_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = max{1, x1 - 7, x2}
            sumwith_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 15}
            cons_2!_A(x1,x2) = max{1, x1 - 23, x2 - 15}
        
  
    partial status:
  
      pi(sumwith_2#) = [2]
      pi(cons_2) = [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: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(plus_0(),y) -> plus_1(y)
r4: app(plus_1(x1),y) -> plus_2(x1,y)
r5: app(s_0(),y) -> s_1(y)
r6: app(sumwith_0(),y) -> sumwith_1(y)
r7: app(sumwith_1(x1),y) -> sumwith_2(x1,y)
r8: plus_2(|0|_0(),y) -> y
r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r10: sumwith_2(f,nil_0()) -> nil_0()
r11: sumwith_2(f,cons_2(x,xs)) -> plus_2(app(f,x),sumwith_2(f,xs))

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:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{3, x1 - 3}
            plus_2#!_A(x1,x2) = max{0, x1 - 20}
            s_1!_A(x1) = max{4, x1 - 19}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{3, x1 - 3}
            plus_2#!_A(x1,x2) = max{3, x1 - 4, x2 - 1}
            s_1!_A(x1) = max{2, x1 - 17}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.