/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

som3(g,a,a)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) PiDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


----------------------------------------

(0)
Obligation:
Clauses:

som3([], Bs, Bs).
som3(As, [], As).
som3(.(A, As), .(B, Bs), .(+(A, B), Cs)) :- som3(As, Bs, Cs).
som4_1(As, Bs, Cs, Ds) :- ','(som3(As, Bs, Es), som3(Es, Cs, Ds)).
som4_2(As, Bs, Cs, Ds) :- ','(som3(Es, Cs, Ds), som3(As, Bs, Es)).


Query: som3(g,a,a)
----------------------------------------

(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

som3_in_3: (b,f,f)

Transforming Prolog into the following Term Rewriting System:

Pi-finite rewrite system:
The TRS R consists of the following rules:

   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)

[]  =  []

som3_out_gaa(x1, x2, x3)  =  som3_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



----------------------------------------

(2)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)

[]  =  []

som3_out_gaa(x1, x2, x3)  =  som3_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)



----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)

[]  =  []

som3_out_gaa(x1, x2, x3)  =  som3_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)

[]  =  []

som3_out_gaa(x1, x2, x3)  =  som3_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
----------------------------------------

(6)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

   som3_in_gaa([], Bs, Bs) -> som3_out_gaa([], Bs, Bs)
   som3_in_gaa(As, [], As) -> som3_out_gaa(As, [], As)
   som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
   U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) -> som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)

[]  =  []

som3_out_gaa(x1, x2, x3)  =  som3_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(7) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) -> SOM3_IN_GAA(As, Bs, Cs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)

SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(9) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SOM3_IN_GAA(.(A, As)) -> SOM3_IN_GAA(As)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(11) QDPSizeChangeProof (EQUIVALENT)
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. 

From the DPs we obtained the following set of size-change graphs:
*SOM3_IN_GAA(.(A, As)) -> SOM3_IN_GAA(As)
The graph contains the following edges 1 > 1


----------------------------------------

(12)
YES