YES

We show the termination of the TRS R:

  concat(leaf(),Y) -> Y
  concat(cons(U,V),Y) -> cons(U,concat(V,Y))
  lessleaves(X,leaf()) -> false()
  lessleaves(leaf(),cons(W,Z)) -> true()
  lessleaves(cons(U,V),cons(W,Z)) -> lessleaves(concat(U,V),concat(W,Z))

-- SCC decomposition.

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

p1: concat#(cons(U,V),Y) -> concat#(V,Y)
p2: lessleaves#(cons(U,V),cons(W,Z)) -> lessleaves#(concat(U,V),concat(W,Z))
p3: lessleaves#(cons(U,V),cons(W,Z)) -> concat#(U,V)
p4: lessleaves#(cons(U,V),cons(W,Z)) -> concat#(W,Z)

and R consists of:

r1: concat(leaf(),Y) -> Y
r2: concat(cons(U,V),Y) -> cons(U,concat(V,Y))
r3: lessleaves(X,leaf()) -> false()
r4: lessleaves(leaf(),cons(W,Z)) -> true()
r5: lessleaves(cons(U,V),cons(W,Z)) -> lessleaves(concat(U,V),concat(W,Z))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: lessleaves#(cons(U,V),cons(W,Z)) -> lessleaves#(concat(U,V),concat(W,Z))

and R consists of:

r1: concat(leaf(),Y) -> Y
r2: concat(cons(U,V),Y) -> cons(U,concat(V,Y))
r3: lessleaves(X,leaf()) -> false()
r4: lessleaves(leaf(),cons(W,Z)) -> true()
r5: lessleaves(cons(U,V),cons(W,Z)) -> lessleaves(concat(U,V),concat(W,Z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            lessleaves#_A(x1,x2) = x1 + x2 + (1,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,4)
            leaf_A() = (1,1)
    
      precedence:
      
        concat = leaf > cons > lessleaves#
      
      partial status:
    
        pi(lessleaves#) = [1, 2]
        pi(cons) = [1, 2]
        pi(concat) = [1]
        pi(leaf) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            lessleaves#_A(x1,x2) = x1 + x2
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,3)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + (2,2)
            leaf_A() = (1,1)
    
      precedence:
      
        lessleaves# = cons = concat = leaf
      
      partial status:
    
        pi(lessleaves#) = [2]
        pi(cons) = [2]
        pi(concat) = []
        pi(leaf) = []
    

The next rules are strictly ordered:

  p1

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: concat#(cons(U,V),Y) -> concat#(V,Y)

and R consists of:

r1: concat(leaf(),Y) -> Y
r2: concat(cons(U,V),Y) -> cons(U,concat(V,Y))
r3: lessleaves(X,leaf()) -> false()
r4: lessleaves(leaf(),cons(W,Z)) -> true()
r5: lessleaves(cons(U,V),cons(W,Z)) -> lessleaves(concat(U,V),concat(W,Z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            concat#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
            cons_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
    
      precedence:
      
        concat# = cons
      
      partial status:
    
        pi(concat#) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            concat#_A(x1,x2) = (1,1)
            cons_A(x1,x2) = x1 + (2,2)
    
      precedence:
      
        concat# = cons
      
      partial status:
    
        pi(concat#) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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