YES
We show the termination of the TRS R:

  app(app(mapbt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
  app(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app(app(app(branch(),app(f,x)),app(app(mapbt(),f),l)),app(app(mapbt(),f),r))

Instead we prove termination of the uncurried system:

  app(branch_0(),y) -> branch_1(y)
  app(branch_1(x1),y) -> branch_2(x1,y)
  app(branch_2(x1,x2),y) -> branch_3(x1,x2,y)
  app(leaf_0(),y) -> leaf_1(y)
  app(mapbt_0(),y) -> mapbt_1(y)
  app(mapbt_1(x1),y) -> mapbt_2(x1,y)
  mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x))
  mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r))


-- SCC decomposition.

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

p1: app#(mapbt_1(x1),y) -> mapbt_2#(x1,y)
p2: mapbt_2#(f,leaf_1(x)) -> app#(f,x)
p3: mapbt_2#(f,branch_3(x,l,r)) -> app#(f,x)
p4: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l)
p5: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r)

and R consists of:

r1: app(branch_0(),y) -> branch_1(y)
r2: app(branch_1(x1),y) -> branch_2(x1,y)
r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y)
r4: app(leaf_0(),y) -> leaf_1(y)
r5: app(mapbt_0(),y) -> mapbt_1(y)
r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y)
r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x))
r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: app#(mapbt_1(x1),y) -> mapbt_2#(x1,y)
p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r)
p3: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l)
p4: mapbt_2#(f,branch_3(x,l,r)) -> app#(f,x)
p5: mapbt_2#(f,leaf_1(x)) -> app#(f,x)

and R consists of:

r1: app(branch_0(),y) -> branch_1(y)
r2: app(branch_1(x1),y) -> branch_2(x1,y)
r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y)
r4: app(leaf_0(),y) -> leaf_1(y)
r5: app(mapbt_0(),y) -> mapbt_1(y)
r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y)
r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x))
r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r))

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:
          
            app#_A(x1,x2) = max{0, x1 - 9, x2 - 19}
            mapbt_1_A(x1) = max{8, x1 + 2}
            mapbt_2#_A(x1,x2) = max{0, x1 - 9, x2 - 19}
            branch_3_A(x1,x2,x3) = max{x1 + 16, x2 + 16, x3 + 17}
            leaf_1_A(x1) = x1 + 18
            app#!_A(x1,x2) = max{6, x1 - 9, x2 - 1}
            mapbt_1!_A(x1) = max{3, x1 - 24}
            mapbt_2#!_A(x1,x2) = max{4, x1 - 8, x2 - 11}
            branch_3!_A(x1,x2,x3) = max{1, x1 - 12, x2 - 32, x3 - 21}
            leaf_1!_A(x1) = max{1, x1 - 8}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 3, x2 - 19}
            mapbt_1_A(x1) = x1 + 2
            mapbt_2#_A(x1,x2) = max{0, x1 - 3, x2 - 35}
            branch_3_A(x1,x2,x3) = max{34, x1 + 16, x2 + 8, x3}
            leaf_1_A(x1) = max{24, x1 + 16}
            app#!_A(x1,x2) = max{6, x1 - 3, x2 - 2}
            mapbt_1!_A(x1) = max{4, x1 - 12}
            mapbt_2#!_A(x1,x2) = max{5, x1 - 2, x2 - 17}
            branch_3!_A(x1,x2,x3) = max{0, x2 - 1, x3 - 9}
            leaf_1!_A(x1) = max{2, x1 - 21}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(mapbt_1) = [1]
      pi(mapbt_2#) = [1]
      pi(branch_3) = [3]
      pi(leaf_1) = [1]

The following rules are removed:

  p1, p4, p5



-- SCC decomposition.

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

p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r)
p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l)

and R consists of:

r1: app(branch_0(),y) -> branch_1(y)
r2: app(branch_1(x1),y) -> branch_2(x1,y)
r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y)
r4: app(leaf_0(),y) -> leaf_1(y)
r5: app(mapbt_0(),y) -> mapbt_1(y)
r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y)
r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x))
r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r)
p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l)

and R consists of:

r1: app(branch_0(),y) -> branch_1(y)
r2: app(branch_1(x1),y) -> branch_2(x1,y)
r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y)
r4: app(leaf_0(),y) -> leaf_1(y)
r5: app(mapbt_0(),y) -> mapbt_1(y)
r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y)
r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x))
r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r))

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:
          
            mapbt_2#_A(x1,x2) = 0
            branch_3_A(x1,x2,x3) = max{3, x1 - 2, x2, x3 - 3}
            mapbt_2#!_A(x1,x2) = x2 + 5
            branch_3!_A(x1,x2,x3) = max{12, x1 - 9, x2 + 9, x3 + 6}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            mapbt_2#_A(x1,x2) = 0
            branch_3_A(x1,x2,x3) = max{3, x1 - 3, x2 - 3, x3 - 5}
            mapbt_2#!_A(x1,x2) = max{5, x1 + 1}
            branch_3!_A(x1,x2,x3) = max{4, x1 - 5, x2 - 15, x3 - 3}
        
  
    partial status:
  
      pi(mapbt_2#) = [2]
      pi(branch_3) = [2, 3]

The following rules are removed:

  p1, p2

Then no dependency pair remains.