85.46/82.31	MAYBE
85.49/82.32	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
85.49/82.32	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
85.49/82.32	
85.49/82.32	
85.49/82.32	Outermost Termination of the given OTRS could not be shown:
85.49/82.32	
85.49/82.32	(0) OTRS
85.49/82.32	(1) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms]
85.49/82.32	(2) QTRS
85.49/82.32	    (3) QTRSRRRProof [EQUIVALENT, 65 ms]
85.49/82.32	    (4) QTRS
85.49/82.32	    (5) DependencyPairsProof [EQUIVALENT, 1 ms]
85.49/82.32	    (6) QDP
85.49/82.32	    (7) DependencyGraphProof [EQUIVALENT, 0 ms]
85.49/82.32	    (8) AND
85.49/82.32	        (9) QDP
85.49/82.32	            (10) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	            (11) QDP
85.49/82.32	            (12) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	            (13) QDP
85.49/82.32	            (14) MRRProof [EQUIVALENT, 0 ms]
85.49/82.32	            (15) QDP
85.49/82.32	            (16) PisEmptyProof [EQUIVALENT, 0 ms]
85.49/82.32	            (17) YES
85.49/82.32	        (18) QDP
85.49/82.32	            (19) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	            (20) QDP
85.49/82.32	                (21) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	                (22) QDP
85.49/82.32	                (23) TransformationProof [SOUND, 0 ms]
85.49/82.32	                (24) QDP
85.49/82.32	                (25) DependencyGraphProof [EQUIVALENT, 0 ms]
85.49/82.32	                (26) QDP
85.49/82.32	            (27) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	            (28) QDP
85.49/82.32	                (29) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	                (30) QDP
85.49/82.32	(31) Trivial-Transformation [SOUND, 0 ms]
85.49/82.32	(32) QTRS
85.49/82.32	    (33) QTRSRRRProof [EQUIVALENT, 35 ms]
85.49/82.32	    (34) QTRS
85.49/82.32	    (35) AAECC Innermost [EQUIVALENT, 0 ms]
85.49/82.32	    (36) QTRS
85.49/82.32	    (37) DependencyPairsProof [EQUIVALENT, 0 ms]
85.49/82.32	    (38) QDP
85.49/82.32	    (39) DependencyGraphProof [EQUIVALENT, 0 ms]
85.49/82.32	    (40) QDP
85.49/82.32	    (41) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	    (42) QDP
85.49/82.32	    (43) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	    (44) QDP
85.49/82.32	    (45) TransformationProof [EQUIVALENT, 0 ms]
85.49/82.32	    (46) QDP
85.49/82.32	    (47) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	    (48) QDP
85.49/82.32	    (49) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	    (50) QDP
85.49/82.32	    (51) TransformationProof [EQUIVALENT, 0 ms]
85.49/82.32	    (52) QDP
85.49/82.32	    (53) UsableRulesProof [EQUIVALENT, 0 ms]
85.49/82.32	    (54) QDP
85.49/82.32	    (55) QReductionProof [EQUIVALENT, 0 ms]
85.49/82.32	    (56) QDP
85.49/82.32	    (57) TransformationProof [EQUIVALENT, 0 ms]
85.49/82.32	    (58) QDP
85.49/82.32	    (59) NonTerminationLoopProof [COMPLETE, 0 ms]
85.49/82.32	    (60) NO
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(0)
85.49/82.32	Obligation:
85.49/82.32	Term rewrite system R:
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   f(x, x) -> f(0, 1)
85.49/82.32	   0 -> 2
85.49/82.32	   1 -> 2
85.49/82.32	   f(x, 2) -> 2
85.49/82.32	   f(2, x) -> 2
85.49/82.32	
85.49/82.32	
85.49/82.32	
85.49/82.32	Outermost Strategy.
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(1) Thiemann-SpecialC-Transformation (EQUIVALENT)
85.49/82.32	We applied the Thiemann-SpecialC transformation  to transform the outermost TRS to an innermost TRS.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(2)
85.49/82.32	Obligation:
85.49/82.32	Q restricted rewrite system:
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   top(go_up(x)) -> top(reduce(x))
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, 2) -> result_f(2)
85.49/82.32	   redex_f(2, x) -> result_f(2)
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(3) QTRSRRRProof (EQUIVALENT)
85.49/82.32	Used ordering:
85.49/82.32	Polynomial interpretation [POLO]:
85.49/82.32	
85.49/82.32	   POL(0) = 0
85.49/82.32	   POL(1) = 0
85.49/82.32	   POL(2) = 0
85.49/82.32	   POL(check_f(x_1)) = x_1
85.49/82.32	   POL(f(x_1, x_2)) = 1 + x_1 + x_2
85.49/82.32	   POL(go_up(x_1)) = 2*x_1
85.49/82.32	   POL(in_f_1(x_1, x_2)) = 2 + x_1 + 2*x_2
85.49/82.32	   POL(in_f_2(x_1, x_2)) = 2 + 2*x_1 + x_2
85.49/82.32	   POL(redex_f(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
85.49/82.32	   POL(reduce(x_1)) = 2*x_1
85.49/82.32	   POL(result_f(x_1)) = 2*x_1
85.49/82.32	   POL(top(x_1)) = x_1
85.49/82.32	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
85.49/82.32	
85.49/82.32	   redex_f(x, 2) -> result_f(2)
85.49/82.32	   redex_f(2, x) -> result_f(2)
85.49/82.32	
85.49/82.32	
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(4)
85.49/82.32	Obligation:
85.49/82.32	Q restricted rewrite system:
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   top(go_up(x)) -> top(reduce(x))
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(5) DependencyPairsProof (EQUIVALENT)
85.49/82.32	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(6)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.32	   TOP(go_up(x)) -> REDUCE(x)
85.49/82.32	   REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2))
85.49/82.32	   REDUCE(f(x_1, x_2)) -> REDEX_F(x_1, x_2)
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> IN_F_1(reduce(x_1), x_2)
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1)
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> IN_F_2(x_1, reduce(x_2))
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2)
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   top(go_up(x)) -> top(reduce(x))
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(7) DependencyGraphProof (EQUIVALENT)
85.49/82.32	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(8)
85.49/82.32	Complex Obligation (AND)
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(9)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1)
85.49/82.32	   REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2))
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2)
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   top(go_up(x)) -> top(reduce(x))
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(10) UsableRulesProof (EQUIVALENT)
85.49/82.32	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(11)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1)
85.49/82.32	   REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2))
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2)
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(12) QReductionProof (EQUIVALENT)
85.49/82.32	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(13)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1)
85.49/82.32	   REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2))
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2)
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(14) MRRProof (EQUIVALENT)
85.49/82.32	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
85.49/82.32	
85.49/82.32	Strictly oriented dependency pairs:
85.49/82.32	
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1)
85.49/82.32	   REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2))
85.49/82.32	   CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2)
85.49/82.32	
85.49/82.32	
85.49/82.32	Used ordering: Polynomial interpretation [POLO]:
85.49/82.32	
85.49/82.32	   POL(0) = 0
85.49/82.32	   POL(1) = 0
85.49/82.32	   POL(CHECK_F(x_1)) = 1 + 2*x_1
85.49/82.32	   POL(REDUCE(x_1)) = 2 + 2*x_1
85.49/82.32	   POL(f(x_1, x_2)) = 1 + x_1 + 2*x_2
85.49/82.32	   POL(redex_f(x_1, x_2)) = 1 + x_1 + x_2
85.49/82.32	   POL(result_f(x_1)) = x_1
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(15)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	P is empty.
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(16) PisEmptyProof (EQUIVALENT)
85.49/82.32	The TRS P is empty. Hence, there is no (P,Q,R) chain.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(17)
85.49/82.32	YES
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(18)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   top(go_up(x)) -> top(reduce(x))
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(19) UsableRulesProof (EQUIVALENT)
85.49/82.32	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(20)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(21) QReductionProof (EQUIVALENT)
85.49/82.32	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(22)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(23) TransformationProof (SOUND)
85.49/82.32	By narrowing [LPAR04] the rule TOP(go_up(x)) -> TOP(reduce(x)) at position [0] we obtained the following new rules [LPAR04]:
85.49/82.32	
85.49/82.32	   (TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1))),TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1))))
85.49/82.32	   (TOP(go_up(0)) -> TOP(go_up(2)),TOP(go_up(0)) -> TOP(go_up(2)))
85.49/82.32	   (TOP(go_up(1)) -> TOP(go_up(2)),TOP(go_up(1)) -> TOP(go_up(2)))
85.49/82.32	
85.49/82.32	
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(24)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1)))
85.49/82.32	   TOP(go_up(0)) -> TOP(go_up(2))
85.49/82.32	   TOP(go_up(1)) -> TOP(go_up(2))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(25) DependencyGraphProof (EQUIVALENT)
85.49/82.32	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(26)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1)))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.32	   in_f_2(x0, go_up(x1))
85.49/82.32	
85.49/82.32	We have to consider all minimal (P,Q,R)-chains.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(27) UsableRulesProof (EQUIVALENT)
85.49/82.32	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.32	----------------------------------------
85.49/82.32	
85.49/82.32	(28)
85.49/82.32	Obligation:
85.49/82.32	Q DP problem:
85.49/82.32	The TRS P consists of the following rules:
85.49/82.32	
85.49/82.32	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.32	
85.49/82.32	The TRS R consists of the following rules:
85.49/82.32	
85.49/82.32	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.32	   reduce(0) -> go_up(2)
85.49/82.32	   reduce(1) -> go_up(2)
85.49/82.32	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.32	   check_f(result_f(x)) -> go_up(x)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.32	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.32	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.32	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.32	
85.49/82.32	The set Q consists of the following terms:
85.49/82.32	
85.49/82.32	   top(go_up(x0))
85.49/82.32	   reduce(f(x0, x1))
85.49/82.32	   redex_f(x0, x0)
85.49/82.32	   reduce(0)
85.49/82.32	   reduce(1)
85.49/82.32	   redex_f(x0, 2)
85.49/82.32	   redex_f(2, x0)
85.49/82.32	   check_f(result_f(x0))
85.49/82.32	   check_f(redex_f(x0, x1))
85.49/82.32	   in_f_1(go_up(x0), x1)
85.49/82.33	   in_f_2(x0, go_up(x1))
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(29) QReductionProof (EQUIVALENT)
85.49/82.33	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.33	
85.49/82.33	   top(go_up(x0))
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(30)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   TOP(go_up(x)) -> TOP(reduce(x))
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2))
85.49/82.33	   reduce(0) -> go_up(2)
85.49/82.33	   reduce(1) -> go_up(2)
85.49/82.33	   redex_f(x, x) -> result_f(f(0, 1))
85.49/82.33	   check_f(result_f(x)) -> go_up(x)
85.49/82.33	   check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2)
85.49/82.33	   check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2))
85.49/82.33	   in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2))
85.49/82.33	   in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2))
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   reduce(f(x0, x1))
85.49/82.33	   redex_f(x0, x0)
85.49/82.33	   reduce(0)
85.49/82.33	   reduce(1)
85.49/82.33	   redex_f(x0, 2)
85.49/82.33	   redex_f(2, x0)
85.49/82.33	   check_f(result_f(x0))
85.49/82.33	   check_f(redex_f(x0, x1))
85.49/82.33	   in_f_1(go_up(x0), x1)
85.49/82.33	   in_f_2(x0, go_up(x1))
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(31) Trivial-Transformation (SOUND)
85.49/82.33	We applied the Trivial transformation  to transform the outermost TRS to a standard TRS.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(32)
85.49/82.33	Obligation:
85.49/82.33	Q restricted rewrite system:
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	   f(x, 2) -> 2
85.49/82.33	   f(2, x) -> 2
85.49/82.33	
85.49/82.33	Q is empty.
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(33) QTRSRRRProof (EQUIVALENT)
85.49/82.33	Used ordering:
85.49/82.33	Polynomial interpretation [POLO]:
85.49/82.33	
85.49/82.33	   POL(0) = 0
85.49/82.33	   POL(1) = 0
85.49/82.33	   POL(2) = 0
85.49/82.33	   POL(f(x_1, x_2)) = 1 + x_1 + x_2
85.49/82.33	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
85.49/82.33	
85.49/82.33	   f(x, 2) -> 2
85.49/82.33	   f(2, x) -> 2
85.49/82.33	
85.49/82.33	
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(34)
85.49/82.33	Obligation:
85.49/82.33	Q restricted rewrite system:
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	Q is empty.
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(35) AAECC Innermost (EQUIVALENT)
85.49/82.33	We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The TRS R 2 is 
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	
85.49/82.33	The signature Sigma is {f_2}
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(36)
85.49/82.33	Obligation:
85.49/82.33	Q restricted rewrite system:
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   f(x0, x0)
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(37) DependencyPairsProof (EQUIVALENT)
85.49/82.33	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(38)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(0, 1)
85.49/82.33	   F(x, x) -> 0^1
85.49/82.33	   F(x, x) -> 1^1
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   f(x0, x0)
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(39) DependencyGraphProof (EQUIVALENT)
85.49/82.33	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(40)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(0, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   f(x, x) -> f(0, 1)
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   f(x0, x0)
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(41) UsableRulesProof (EQUIVALENT)
85.49/82.33	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(42)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(0, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   f(x0, x0)
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(43) QReductionProof (EQUIVALENT)
85.49/82.33	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.33	
85.49/82.33	   f(x0, x0)
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(44)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(0, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(45) TransformationProof (EQUIVALENT)
85.49/82.33	By rewriting [LPAR04] the rule F(x, x) -> F(0, 1) at position [0] we obtained the following new rules [LPAR04]:
85.49/82.33	
85.49/82.33	   (F(x, x) -> F(2, 1),F(x, x) -> F(2, 1))
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(46)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   0 -> 2
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(47) UsableRulesProof (EQUIVALENT)
85.49/82.33	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(48)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   0
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(49) QReductionProof (EQUIVALENT)
85.49/82.33	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.33	
85.49/82.33	   0
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(50)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 1)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(51) TransformationProof (EQUIVALENT)
85.49/82.33	By rewriting [LPAR04] the rule F(x, x) -> F(2, 1) at position [1] we obtained the following new rules [LPAR04]:
85.49/82.33	
85.49/82.33	   (F(x, x) -> F(2, 2),F(x, x) -> F(2, 2))
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(52)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 2)
85.49/82.33	
85.49/82.33	The TRS R consists of the following rules:
85.49/82.33	
85.49/82.33	   1 -> 2
85.49/82.33	
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(53) UsableRulesProof (EQUIVALENT)
85.49/82.33	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(54)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 2)
85.49/82.33	
85.49/82.33	R is empty.
85.49/82.33	The set Q consists of the following terms:
85.49/82.33	
85.49/82.33	   1
85.49/82.33	
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(55) QReductionProof (EQUIVALENT)
85.49/82.33	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
85.49/82.33	
85.49/82.33	   1
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(56)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(x, x) -> F(2, 2)
85.49/82.33	
85.49/82.33	R is empty.
85.49/82.33	Q is empty.
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(57) TransformationProof (EQUIVALENT)
85.49/82.33	By instantiating [LPAR04] the rule F(x, x) -> F(2, 2) we obtained the following new rules [LPAR04]:
85.49/82.33	
85.49/82.33	   (F(2, 2) -> F(2, 2),F(2, 2) -> F(2, 2))
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(58)
85.49/82.33	Obligation:
85.49/82.33	Q DP problem:
85.49/82.33	The TRS P consists of the following rules:
85.49/82.33	
85.49/82.33	   F(2, 2) -> F(2, 2)
85.49/82.33	
85.49/82.33	R is empty.
85.49/82.33	Q is empty.
85.49/82.33	We have to consider all minimal (P,Q,R)-chains.
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(59) NonTerminationLoopProof (COMPLETE)
85.49/82.33	We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
85.49/82.33	Found a loop by semiunifying a rule from P directly.
85.49/82.33	
85.49/82.33	s = F(2, 2) evaluates to  t =F(2, 2)
85.49/82.33	
85.49/82.33	Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
85.49/82.33	* Matcher: [ ]
85.49/82.33	* Semiunifier: [ ]
85.49/82.33	
85.49/82.33	--------------------------------------------------------------------------------
85.49/82.33	Rewriting sequence
85.49/82.33	
85.49/82.33	The DP semiunifies directly so there is only one rewrite step from F(2, 2) to F(2, 2).
85.49/82.33	
85.49/82.33	
85.49/82.33	
85.49/82.33	
85.49/82.33	----------------------------------------
85.49/82.33	
85.49/82.33	(60)
85.49/82.33	NO
85.64/82.41	EOF