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(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(app(curry(),g),x),y) -> app(app(g,x),y)
  inc() -> app(map(),app(app(curry(),plus()),app(s(),|0|())))

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(curry_0(),y) -> curry_1(y)
  app(curry_1(x1),y) -> curry_2(x1,y)
  app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_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)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  curry_3(g,x,y) -> app(app(g,x),y)
  inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))


-- SCC decomposition.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(plus_1(x1),y) -> plus_2#(x1,y)
p4: plus_2#(s_1(x),y) -> plus_2#(x,y)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p7: curry_3#(g,x,y) -> app#(app(g,x),y)
p8: curry_3#(g,x,y) -> app#(g,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p6, p7, p8}
  {p4}


-- Reduction pair.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(g,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: curry_3#(g,x,y) -> app#(app(g,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(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

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:
          app#_A(x1,x2) = x1 + 2
          curry_2_A(x1,x2) = x1 + x2 + 7
          curry_3#_A(x1,x2,x3) = x1 + x2 + 6
          map_1_A(x1) = x1 + 4
          map_2#_A(x1,x2) = x1 + 3
          cons_2_A(x1,x2) = 1
          app_A(x1,x2) = x1 + x2 + 3
          cons_0_A() = 1
          cons_1_A(x1) = x1 + 2
          curry_0_A() = 3
          curry_1_A(x1) = x1 + 5
          curry_3_A(x1,x2,x3) = x1 + x2 + x3 + 8
          map_0_A() = 2
          map_2_A(x1,x2) = 2
          plus_0_A() = 1
          plus_1_A(x1) = x1 + 2
          plus_2_A(x1,x2) = x1 + x2 + 3
          s_0_A() = 2
          s_1_A(x1) = 1
          |0|_0_A() = 1
          nil_0_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = x1
            curry_2_A(x1,x2) = x1 + x2 + 2
            curry_3#_A(x1,x2,x3) = x1 + x2 + 2
            map_1_A(x1) = x1 + 1
            map_2#_A(x1,x2) = x1
            cons_2_A(x1,x2) = 1
            app_A(x1,x2) = x1 + x2 + 2
            cons_0_A() = 1
            cons_1_A(x1) = x1 + 1
            curry_0_A() = 1
            curry_1_A(x1) = x1 + 1
            curry_3_A(x1,x2,x3) = x1 + x2 + x3 + 4
            map_0_A() = 1
            map_2_A(x1,x2) = 1
            plus_0_A() = 1
            plus_1_A(x1) = x1 + 1
            plus_2_A(x1,x2) = x2 + 1
            s_0_A() = 1
            s_1_A(x1) = 1
            |0|_0_A() = 1
            nil_0_A() = 1
            app#!_A(x1,x2) = 4
            curry_2!_A(x1,x2) = 2
            curry_3#!_A(x1,x2,x3) = x1 + x2 + 5
            map_1!_A(x1) = 3
            map_2#!_A(x1,x2) = x1 + 5
            cons_2!_A(x1,x2) = x1 + 1
            app!_A(x1,x2) = x2 + 4
            cons_0!_A() = 1
            cons_1!_A(x1) = 0
            curry_0!_A() = 0
            curry_1!_A(x1) = 1
            curry_3!_A(x1,x2,x3) = 3
            map_0!_A() = 0
            map_2!_A(x1,x2) = 2
            plus_0!_A() = 8
            plus_1!_A(x1) = 7
            plus_2!_A(x1,x2) = x1 + 6
            s_0!_A() = 6
            s_1!_A(x1) = 5
            |0|_0!_A() = 0
            nil_0!_A() = 2
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = [1, 2]
      pi(curry_3#) = [1]
      pi(map_1) = []
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app) = [1]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_3) = [2]
      pi(map_0) = []
      pi(map_2) = []
      pi(plus_0) = []
      pi(plus_1) = [1]
      pi(plus_2) = [1]
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_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(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_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(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The set of usable rules consists of

  (no rules)

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:
          map_2#_A(x1,x2) = x2
          cons_2_A(x1,x2) = x1 + x2 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            map_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = x2 + 2
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = x2
  
    partial status:
  
      pi(map_2#) = []
      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: 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(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The set of usable rules consists of

  (no rules)

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:
          plus_2#_A(x1,x2) = x1
          s_1_A(x1) = x1 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            plus_2#_A(x1,x2) = x1
            s_1_A(x1) = x1 + 2
            plus_2#!_A(x1,x2) = 6
            s_1!_A(x1) = 5
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.