2.89/1.32	YES
2.89/1.32	
2.89/1.32	Proof:
2.89/1.32	This system is confluent.
2.89/1.32	By \cite{ALS94}, Theorem 4.1.
2.89/1.32	This system is of type 3 or smaller.
2.89/1.32	This system is strongly deterministic.
2.89/1.32	This system is quasi-decreasing.
2.89/1.32	By \cite{A14}, Theorem 11.5.9.
2.89/1.32	This system is of type 3 or smaller.
2.89/1.32	This system is deterministic.
2.89/1.32	System R transformed to V(R) + Emb.
2.89/1.32	This system is terminating.
2.89/1.32	Call external tool:
2.89/1.32	./ttt2.sh
2.89/1.33	Input:
2.89/1.33	  f(g(x)) -> b
2.89/1.33	  f(g(x)) -> x
2.89/1.33	  g(x) -> c
2.89/1.33	  g(x) -> x
2.89/1.33	  g(x) -> x
2.89/1.33	  f(x) -> x
2.89/1.33	
2.89/1.33	 Matrix Interpretation Processor: dim=3
2.89/1.33	  
2.89/1.33	  interpretation:
2.89/1.33	         [0]
2.89/1.33	   [c] = [0]
2.89/1.33	         [0],
2.89/1.33	   
2.89/1.33	         [0]
2.89/1.33	   [b] = [0]
2.89/1.33	         [1],
2.89/1.33	   
2.89/1.33	             [1 0 0]  
2.89/1.33	   [f](x0) = [0 1 0]x0
2.89/1.33	             [1 0 1]  ,
2.89/1.33	   
2.89/1.33	                  [1]
2.89/1.33	   [g](x0) = x0 + [0]
2.89/1.33	                  [0]
2.89/1.33	  orientation:
2.89/1.33	             [1 0 0]    [1]    [0]      
2.89/1.33	   f(g(x)) = [0 1 0]x + [0] >= [0] = b()
2.89/1.33	             [1 0 1]    [1]    [1]      
2.89/1.33	   
2.89/1.33	             [1 0 0]    [1]         
2.89/1.33	   f(g(x)) = [0 1 0]x + [0] >= x = x
2.89/1.33	             [1 0 1]    [1]         
2.89/1.33	   
2.89/1.33	              [1]    [0]      
2.89/1.33	   g(x) = x + [0] >= [0] = c()
2.89/1.33	              [0]    [0]      
2.89/1.33	   
2.89/1.33	              [1]         
2.89/1.33	   g(x) = x + [0] >= x = x
2.89/1.33	              [0]         
2.89/1.33	   
2.89/1.33	          [1 0 0]          
2.89/1.33	   f(x) = [0 1 0]x >= x = x
2.89/1.33	          [1 0 1]          
2.89/1.33	  problem:
2.89/1.33	   f(x) -> x
2.89/1.33	  Matrix Interpretation Processor: dim=3
2.89/1.33	   
2.89/1.33	   interpretation:
2.89/1.33	                   [1]
2.89/1.33	    [f](x0) = x0 + [0]
2.89/1.33	                   [0]
2.89/1.33	   orientation:
2.89/1.33	               [1]         
2.89/1.33	    f(x) = x + [0] >= x = x
2.89/1.33	               [0]         
2.89/1.33	   problem:
2.89/1.33	    
2.89/1.33	   Qed
2.89/1.33	All 1 critical pairs are joinable.
2.89/1.33	Overlap: (rule1: f(g(y)) -> b <= y = a, rule2: g(z) -> c <= z = c, pos: 1, mgu: {(z,y)})
2.89/1.33	CP: f(c) = b <= y = a, y = c
2.89/1.33	This critical pair is unfeasible.
2.89/1.33	
3.23/1.72	EOF