0.00/0.01	YES
0.00/0.01	We consider the system theBenchmark.
0.00/0.01	
0.00/0.01	  Alphabet:
0.00/0.01	
0.00/0.01	    cons : [t * f] --> f 
0.00/0.01	    heightf : [f] --> N 
0.00/0.01	    heightt : [t] --> N 
0.00/0.01	    leaf : [] --> t 
0.00/0.01	    max : [N * N] --> N 
0.00/0.01	    nil : [] --> f 
0.00/0.01	    node : [f] --> t 
0.00/0.01	    s : [N] --> N 
0.00/0.01	    z : [] --> N 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    heightf(nil) => z 
0.00/0.01	    heightf(cons(x, y)) => max(heightt(x), heightf(y)) 
0.00/0.01	    heightt(leaf) => z 
0.00/0.01	    heightt(node(x)) => s(heightf(x)) 
0.00/0.01	
0.00/0.01	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.01	
0.00/0.01	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.01	
0.00/0.01	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.01	
0.00/0.01	  heightf(nil) >? z 
0.00/0.01	  heightf(cons(X, Y)) >? max(heightt(X), heightf(Y)) 
0.00/0.01	  heightt(leaf) >? z 
0.00/0.01	  heightt(node(X)) >? s(heightf(X)) 
0.00/0.01	
0.00/0.01	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.01	
0.00/0.01	The following interpretation satisfies the requirements:
0.00/0.01	
0.00/0.01	  cons = \y0y1.3 + 3y0 + 3y1 
0.00/0.01	  heightf = \y0.y0 
0.00/0.01	  heightt = \y0.y0 
0.00/0.01	  leaf = 3 
0.00/0.01	  max = \y0y1.y0 + y1 
0.00/0.01	  nil = 3 
0.00/0.01	  node = \y0.3 + 3y0 
0.00/0.01	  s = \y0.y0 
0.00/0.01	  z = 0 
0.00/0.01	
0.00/0.01	Using this interpretation, the requirements translate to:
0.00/0.01	
0.00/0.01	  [[heightf(nil)]] = 3 > 0 = [[z]] 
0.00/0.01	  [[heightf(cons(_x0, _x1))]] = 3 + 3x0 + 3x1 > x0 + x1 = [[max(heightt(_x0), heightf(_x1))]] 
0.00/0.01	  [[heightt(leaf)]] = 3 > 0 = [[z]] 
0.00/0.01	  [[heightt(node(_x0))]] = 3 + 3x0 > x0 = [[s(heightf(_x0))]] 
0.00/0.01	
0.00/0.01	We can thus remove the following rules:
0.00/0.01	
0.00/0.01	  heightf(nil) => z 
0.00/0.01	  heightf(cons(X, Y)) => max(heightt(X), heightf(Y)) 
0.00/0.01	  heightt(leaf) => z 
0.00/0.01	  heightt(node(X)) => s(heightf(X)) 
0.00/0.01	
0.00/0.01	All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.
0.00/0.01	
0.00/0.01	
0.00/0.01	+++ Citations +++
0.00/0.01	
0.00/0.01	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.01	EOF