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))

-- SCC decomposition.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(s(),app(app(plus(),x),y))
p2: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)
p3: app#(app(plus(),app(s(),x)),y) -> app#(plus(),x)
p4: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(app(plus(),app(f,x)),app(app(sumwith(),f),xs))
p5: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(plus(),app(f,x))
p6: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(f,x)
p7: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(app(sumwith(),f),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(sumwith(),f),nil()) -> nil()
r4: app(app(sumwith(),f),app(app(cons(),x),xs)) -> app(app(plus(),app(f,x)),app(app(sumwith(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p6, p7}
  {p2}


-- Reduction pair.

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

p1: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(app(sumwith(),f),app(app(cons(),x),xs)) -> app#(app(sumwith(),f),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(sumwith(),f),nil()) -> nil()
r4: app(app(sumwith(),f),app(app(cons(),x),xs)) -> app(app(plus(),app(f,x)),app(app(sumwith(),f),xs))

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:
        
          app#_A(x1,x2) = max{x1 + 10, x2 + 6}
          app_A(x1,x2) = max{x1 + 3, x2 + 6}
          sumwith_A = 3
          cons_A = 4
    
      precedence:
      
        app# = app = sumwith = cons
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(sumwith) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x2 - 4}
          app_A(x1,x2) = max{x1 + 6, x2 + 3}
          sumwith_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = sumwith = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(sumwith) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1, p2

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: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(sumwith(),f),nil()) -> nil()
r4: app(app(sumwith(),f),app(app(cons(),x),xs)) -> app(app(plus(),app(f,x)),app(app(sumwith(),f),xs))

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:
        
          app#_A(x1,x2) = max{x1 + 2, x2 + 3}
          app_A(x1,x2) = max{x1, x2 + 3}
          plus_A = 2
          s_A = 3
    
      precedence:
      
        app# = app = plus = s
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(plus) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 - 9, x2 + 3}
          app_A(x1,x2) = max{x1 + 2, x2 + 4}
          plus_A = 2
          s_A = 6
    
      precedence:
      
        app# = app = plus = s
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(plus) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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