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))

-- SCC decomposition.

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

p1: app#(app(mapbt(),f),app(leaf(),x)) -> app#(leaf(),app(f,x))
p2: app#(app(mapbt(),f),app(leaf(),x)) -> app#(f,x)
p3: 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))
p4: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(app(branch(),app(f,x)),app(app(mapbt(),f),l))
p5: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(branch(),app(f,x))
p6: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(f,x)
p7: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(app(mapbt(),f),l)
p8: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(app(mapbt(),f),r)

and R consists of:

r1: app(app(mapbt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: 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))

The estimated dependency graph contains the following SCCs:

  {p2, p6, p7, p8}


-- Reduction pair.

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

p1: app#(app(mapbt(),f),app(leaf(),x)) -> app#(f,x)
p2: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(app(mapbt(),f),r)
p3: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(app(mapbt(),f),l)
p4: app#(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app#(f,x)

and R consists of:

r1: app(app(mapbt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: 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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x2
      app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x2
      mapbt_A() = (1,1,1,1)
      leaf_A() = (1,1,1,1)
      branch_A() = (1,1,1,0)

The next rules are strictly ordered:

  p1, p2, p3, p4

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