(0) Obligation:

Clauses:

select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).

Query: select(a,g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

(2) Obligation:

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

selectA_in_aga(T6, .(T6, T7), T7) → selectA_out_aga(T6, .(T6, T7), T7)
selectA_in_aga(T26, .(T13, .(T26, T27)), .(T13, T27)) → selectA_out_aga(T26, .(T13, .(T26, T27)), .(T13, T27))
selectA_in_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_aga(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
selectA_in_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_aga(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))
U2_aga(T78, T51, T75, T76, T79, selectA_out_aga(T78, T76, T79)) → selectA_out_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79)))
U1_aga(T40, T13, T37, T38, T41, selectA_out_aga(T40, T38, T41)) → selectA_out_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41)))

The argument filtering Pi contains the following mapping:
selectA_in_aga(x1, x2, x3)  =  selectA_in_aga(x2)
.(x1, x2)  =  .(x1, x2)
selectA_out_aga(x1, x2, x3)  =  selectA_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4, x5, x6)  =  U1_aga(x2, x3, x4, x6)
U2_aga(x1, x2, x3, x4, x5, x6)  =  U2_aga(x2, x3, x4, x6)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_AGA(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → SELECTA_IN_AGA(T40, T38, T41)
SELECTA_IN_AGA(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_AGA(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))

The TRS R consists of the following rules:

selectA_in_aga(T6, .(T6, T7), T7) → selectA_out_aga(T6, .(T6, T7), T7)
selectA_in_aga(T26, .(T13, .(T26, T27)), .(T13, T27)) → selectA_out_aga(T26, .(T13, .(T26, T27)), .(T13, T27))
selectA_in_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_aga(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
selectA_in_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_aga(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))
U2_aga(T78, T51, T75, T76, T79, selectA_out_aga(T78, T76, T79)) → selectA_out_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79)))
U1_aga(T40, T13, T37, T38, T41, selectA_out_aga(T40, T38, T41)) → selectA_out_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41)))

The argument filtering Pi contains the following mapping:
selectA_in_aga(x1, x2, x3)  =  selectA_in_aga(x2)
.(x1, x2)  =  .(x1, x2)
selectA_out_aga(x1, x2, x3)  =  selectA_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4, x5, x6)  =  U1_aga(x2, x3, x4, x6)
U2_aga(x1, x2, x3, x4, x5, x6)  =  U2_aga(x2, x3, x4, x6)
SELECTA_IN_AGA(x1, x2, x3)  =  SELECTA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5, x6)  =  U1_AGA(x2, x3, x4, x6)
U2_AGA(x1, x2, x3, x4, x5, x6)  =  U2_AGA(x2, x3, x4, x6)

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

(4) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_AGA(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → SELECTA_IN_AGA(T40, T38, T41)
SELECTA_IN_AGA(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_AGA(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))

The TRS R consists of the following rules:

selectA_in_aga(T6, .(T6, T7), T7) → selectA_out_aga(T6, .(T6, T7), T7)
selectA_in_aga(T26, .(T13, .(T26, T27)), .(T13, T27)) → selectA_out_aga(T26, .(T13, .(T26, T27)), .(T13, T27))
selectA_in_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_aga(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
selectA_in_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_aga(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))
U2_aga(T78, T51, T75, T76, T79, selectA_out_aga(T78, T76, T79)) → selectA_out_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79)))
U1_aga(T40, T13, T37, T38, T41, selectA_out_aga(T40, T38, T41)) → selectA_out_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41)))

The argument filtering Pi contains the following mapping:
selectA_in_aga(x1, x2, x3)  =  selectA_in_aga(x2)
.(x1, x2)  =  .(x1, x2)
selectA_out_aga(x1, x2, x3)  =  selectA_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4, x5, x6)  =  U1_aga(x2, x3, x4, x6)
U2_aga(x1, x2, x3, x4, x5, x6)  =  U2_aga(x2, x3, x4, x6)
SELECTA_IN_AGA(x1, x2, x3)  =  SELECTA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5, x6)  =  U1_AGA(x2, x3, x4, x6)
U2_AGA(x1, x2, x3, x4, x5, x6)  =  U2_AGA(x2, x3, x4, x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → SELECTA_IN_AGA(T40, T38, T41)

The TRS R consists of the following rules:

selectA_in_aga(T6, .(T6, T7), T7) → selectA_out_aga(T6, .(T6, T7), T7)
selectA_in_aga(T26, .(T13, .(T26, T27)), .(T13, T27)) → selectA_out_aga(T26, .(T13, .(T26, T27)), .(T13, T27))
selectA_in_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → U1_aga(T40, T13, T37, T38, T41, selectA_in_aga(T40, T38, T41))
selectA_in_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79))) → U2_aga(T78, T51, T75, T76, T79, selectA_in_aga(T78, T76, T79))
U2_aga(T78, T51, T75, T76, T79, selectA_out_aga(T78, T76, T79)) → selectA_out_aga(T78, .(T51, .(T75, T76)), .(T51, .(T75, T79)))
U1_aga(T40, T13, T37, T38, T41, selectA_out_aga(T40, T38, T41)) → selectA_out_aga(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41)))

The argument filtering Pi contains the following mapping:
selectA_in_aga(x1, x2, x3)  =  selectA_in_aga(x2)
.(x1, x2)  =  .(x1, x2)
selectA_out_aga(x1, x2, x3)  =  selectA_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4, x5, x6)  =  U1_aga(x2, x3, x4, x6)
U2_aga(x1, x2, x3, x4, x5, x6)  =  U2_aga(x2, x3, x4, x6)
SELECTA_IN_AGA(x1, x2, x3)  =  SELECTA_IN_AGA(x2)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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:

SELECTA_IN_AGA(T40, .(T13, .(T37, T38)), .(T13, .(T37, T41))) → SELECTA_IN_AGA(T40, T38, T41)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
SELECTA_IN_AGA(x1, x2, x3)  =  SELECTA_IN_AGA(x2)

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

(9) PiDPToQDPProof (SOUND transformation)

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:

SELECTA_IN_AGA(.(T13, .(T37, T38))) → SELECTA_IN_AGA(T38)

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

(11) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • SELECTA_IN_AGA(.(T13, .(T37, T38))) → SELECTA_IN_AGA(T38)
    The graph contains the following edges 1 > 1

(12) YES