3.66/1.73	YES
3.66/1.75	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.66/1.75	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.66/1.75	
3.66/1.75	
3.66/1.75	Left Termination of the query pattern
3.66/1.75	
3.66/1.75	som3(g,a,a)
3.66/1.75	
3.66/1.75	w.r.t. the given Prolog program could successfully be proven:
3.66/1.75	
3.66/1.75	(0) Prolog
3.66/1.75	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.66/1.75	(2) PiTRS
3.66/1.75	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.66/1.75	(4) PiDP
3.66/1.75	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.66/1.75	(6) PiDP
3.66/1.75	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.66/1.75	(8) PiDP
3.66/1.75	(9) PiDPToQDPProof [SOUND, 0 ms]
3.66/1.75	(10) QDP
3.66/1.75	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.66/1.75	(12) YES
3.66/1.75	
3.66/1.75	
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(0)
3.66/1.75	Obligation:
3.66/1.75	Clauses:
3.66/1.75	
3.66/1.75	som3([], Bs, Bs).
3.66/1.75	som3(As, [], As).
3.66/1.75	som3(.(A, As), .(B, Bs), .(+(A, B), Cs)) :- som3(As, Bs, Cs).
3.66/1.75	som4_1(As, Bs, Cs, Ds) :- ','(som3(As, Bs, Es), som3(Es, Cs, Ds)).
3.66/1.75	som4_2(As, Bs, Cs, Ds) :- ','(som3(Es, Cs, Ds), som3(As, Bs, Es)).
3.66/1.75	
3.66/1.75	
3.66/1.75	Query: som3(g,a,a)
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(1) PrologToPiTRSProof (SOUND)
3.66/1.75	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.66/1.75	
3.66/1.75	som3_in_3: (b,f,f)
3.66/1.75	
3.66/1.75	Transforming Prolog into the following Term Rewriting System:
3.66/1.75	
3.66/1.75	Pi-finite rewrite system:
3.66/1.75	The TRS R consists of the following rules:
3.66/1.75	
3.66/1.75	   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
3.66/1.75	   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
3.66/1.75	   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))
3.66/1.75	
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
3.66/1.75	
3.66/1.75	[]  =  []
3.66/1.75	
3.66/1.75	som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
3.66/1.75	
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
3.66/1.75	
3.66/1.75	
3.66/1.75	
3.66/1.75	
3.66/1.75	
3.66/1.75	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.66/1.75	
3.66/1.75	
3.66/1.75	
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(2)
3.66/1.75	Obligation:
3.66/1.75	Pi-finite rewrite system:
3.66/1.75	The TRS R consists of the following rules:
3.66/1.75	
3.66/1.75	   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
3.66/1.75	   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
3.66/1.75	   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))
3.66/1.75	
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
3.66/1.75	
3.66/1.75	[]  =  []
3.66/1.75	
3.66/1.75	som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
3.66/1.75	
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
3.66/1.75	
3.66/1.75	
3.66/1.75	
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(3) DependencyPairsProof (EQUIVALENT)
3.66/1.75	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.66/1.75	Pi DP problem:
3.66/1.75	The TRS P consists of the following rules:
3.66/1.75	
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)
3.66/1.75	
3.66/1.75	The TRS R consists of the following rules:
3.66/1.75	
3.66/1.75	   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
3.66/1.75	   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
3.66/1.75	   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))
3.66/1.75	
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
3.66/1.75	
3.66/1.75	[]  =  []
3.66/1.75	
3.66/1.75	som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
3.66/1.75	
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
3.66/1.75	
3.66/1.75	SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
3.66/1.75	
3.66/1.75	U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)
3.66/1.75	
3.66/1.75	
3.66/1.75	We have to consider all (P,R,Pi)-chains
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(4)
3.66/1.75	Obligation:
3.66/1.75	Pi DP problem:
3.66/1.75	The TRS P consists of the following rules:
3.66/1.75	
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)
3.66/1.75	
3.66/1.75	The TRS R consists of the following rules:
3.66/1.75	
3.66/1.75	   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
3.66/1.75	   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
3.66/1.75	   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))
3.66/1.75	
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
3.66/1.75	
3.66/1.75	[]  =  []
3.66/1.75	
3.66/1.75	som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
3.66/1.75	
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
3.66/1.75	
3.66/1.75	SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
3.66/1.75	
3.66/1.75	U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)
3.66/1.75	
3.66/1.75	
3.66/1.75	We have to consider all (P,R,Pi)-chains
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(5) DependencyGraphProof (EQUIVALENT)
3.66/1.75	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(6)
3.66/1.75	Obligation:
3.66/1.75	Pi DP problem:
3.66/1.75	The TRS P consists of the following rules:
3.66/1.75	
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)
3.66/1.75	
3.66/1.75	The TRS R consists of the following rules:
3.66/1.75	
3.66/1.75	   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
3.66/1.75	   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
3.66/1.75	   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
3.66/1.75	   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))
3.66/1.75	
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
3.66/1.75	
3.66/1.75	[]  =  []
3.66/1.75	
3.66/1.75	som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
3.66/1.75	
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
3.66/1.75	
3.66/1.75	SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
3.66/1.75	
3.66/1.75	
3.66/1.75	We have to consider all (P,R,Pi)-chains
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(7) UsableRulesProof (EQUIVALENT)
3.66/1.75	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(8)
3.66/1.75	Obligation:
3.66/1.75	Pi DP problem:
3.66/1.75	The TRS P consists of the following rules:
3.66/1.75	
3.66/1.75	   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)
3.66/1.75	
3.66/1.75	R is empty.
3.66/1.75	The argument filtering Pi contains the following mapping:
3.66/1.75	.(x1, x2)  =  .(x1, x2)
3.66/1.75	
3.66/1.75	SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
3.66/1.75	
3.66/1.75	
3.66/1.75	We have to consider all (P,R,Pi)-chains
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(9) PiDPToQDPProof (SOUND)
3.66/1.75	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(10)
3.66/1.75	Obligation:
3.66/1.75	Q DP problem:
3.66/1.75	The TRS P consists of the following rules:
3.66/1.75	
3.66/1.75	   SOM3_IN_GAA(.(A, As)) -> SOM3_IN_GAA(As)
3.66/1.75	
3.66/1.75	R is empty.
3.66/1.75	Q is empty.
3.66/1.75	We have to consider all (P,Q,R)-chains.
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(11) QDPSizeChangeProof (EQUIVALENT)
3.66/1.75	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.66/1.75	
3.66/1.75	From the DPs we obtained the following set of size-change graphs:
3.66/1.75	*SOM3_IN_GAA(.(A, As)) -> SOM3_IN_GAA(As)
3.66/1.75	The graph contains the following edges 1 > 1
3.66/1.75	
3.66/1.75	
3.66/1.75	----------------------------------------
3.66/1.75	
3.66/1.75	(12)
3.66/1.75	YES
3.66/1.78	EOF