Left Termination of the query pattern
perm(g,a)
w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToPiTRSProof (⇒, 0 ms)
↳2 PiTRS
↳3 DependencyPairsProof (⇔, 8 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔, 1 ms)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇒, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 PiDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇒, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 PiDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 PiDP
        ↳24 PiDPToQDPProof (⇒, 0 ms)
        ↳25 QDP
        ↳26 MRRProof (⇔, 54 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES

(0) Obligation:

Clauses:

ap1(nil, X, X).
ap1(cons(H, X), Y, cons(H, Z)) :- ap1(X, Y, Z).
ap2(nil, X, X).
ap2(cons(H, X), Y, cons(H, Z)) :- ap2(X, Y, Z).
perm(nil, nil).
perm(Xs, cons(X, Ys)) :- ','(ap1(X1s, cons(X, X2s), Xs), ','(ap2(X1s, X2s, Zs), perm(Zs, Ys))).

Query: perm(g,a)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
perm_in: (b,f)
ap1_in: (f,f,b)
ap2_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)

(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:

PERM_IN_GA(Xs, cons(X, Ys)) → U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
PERM_IN_GA(Xs, cons(X, Ys)) → AP1_IN_AAG(X1s, cons(X, X2s), Xs)
AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z))
AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → AP1_IN_AAG(X, Y, Z)
U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → AP2_IN_GGA(X1s, X2s, Zs)
AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z))
AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → AP2_IN_GGA(X, Y, Z)
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys))
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → PERM_IN_GA(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x5)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3, x4, x5)  =  U2_GGA(x5)
U5_GA(x1, x2, x3, x4)  =  U5_GA(x4)

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

(4) Obligation:

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

PERM_IN_GA(Xs, cons(X, Ys)) → U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
PERM_IN_GA(Xs, cons(X, Ys)) → AP1_IN_AAG(X1s, cons(X, X2s), Xs)
AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z))
AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → AP1_IN_AAG(X, Y, Z)
U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → AP2_IN_GGA(X1s, X2s, Zs)
AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z))
AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → AP2_IN_GGA(X, Y, Z)
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys))
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → PERM_IN_GA(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x5)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3, x4, x5)  =  U2_GGA(x5)
U5_GA(x1, x2, x3, x4)  =  U5_GA(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → AP2_IN_GGA(X, Y, Z)

The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) → AP2_IN_GGA(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x2)
AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

AP2_IN_GGA(cons(X), Y) → AP2_IN_GGA(X, Y)

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

(12) 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:

  • AP2_IN_GGA(cons(X), Y) → AP2_IN_GGA(X, Y)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) YES

(14) Obligation:

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

AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → AP1_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) → AP1_IN_AAG(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x2)
AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)

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

(17) PiDPToQDPProof (SOUND transformation)

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

(18) Obligation:

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

AP1_IN_AAG(cons(Z)) → AP1_IN_AAG(Z)

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

(19) 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:

  • AP1_IN_AAG(cons(Z)) → AP1_IN_AAG(Z)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → PERM_IN_GA(Zs, Ys)
PERM_IN_GA(Xs, cons(X, Ys)) → U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))

The TRS R consists of the following rules:

perm_in_ga(nil, nil) → perm_out_ga(nil, nil)
perm_in_ga(Xs, cons(X, Ys)) → U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))
U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) → perm_out_ga(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ga(x1, x2)  =  perm_in_ga(x1)
nil  =  nil
perm_out_ga(x1, x2)  =  perm_out_ga(x2)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)

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

(22) UsableRulesProof (EQUIVALENT transformation)

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

(23) Obligation:

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

U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) → U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) → PERM_IN_GA(Zs, Ys)
PERM_IN_GA(Xs, cons(X, Ys)) → U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))

The TRS R consists of the following rules:

ap2_in_gga(nil, X, X) → ap2_out_gga(nil, X, X)
ap2_in_gga(cons(H, X), Y, cons(H, Z)) → U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
ap1_in_aag(nil, X, X) → ap1_out_aag(nil, X, X)
ap1_in_aag(cons(H, X), Y, cons(H, Z)) → U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) → ap2_out_gga(cons(H, X), Y, cons(H, Z))
U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) → ap1_out_aag(cons(H, X), Y, cons(H, Z))

The argument filtering Pi contains the following mapping:
nil  =  nil
ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
cons(x1, x2)  =  cons(x2)
ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)

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

(24) PiDPToQDPProof (SOUND transformation)

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

(25) Obligation:

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

U3_GA(ap1_out_aag(X1s, cons(X2s))) → U4_GA(ap2_in_gga(X1s, X2s))
U4_GA(ap2_out_gga(Zs)) → PERM_IN_GA(Zs)
PERM_IN_GA(Xs) → U3_GA(ap1_in_aag(Xs))

The TRS R consists of the following rules:

ap2_in_gga(nil, X) → ap2_out_gga(X)
ap2_in_gga(cons(X), Y) → U2_gga(ap2_in_gga(X, Y))
ap1_in_aag(X) → ap1_out_aag(nil, X)
ap1_in_aag(cons(Z)) → U1_aag(ap1_in_aag(Z))
U2_gga(ap2_out_gga(Z)) → ap2_out_gga(cons(Z))
U1_aag(ap1_out_aag(X, Y)) → ap1_out_aag(cons(X), Y)

The set Q consists of the following terms:

ap2_in_gga(x0, x1)
ap1_in_aag(x0)
U2_gga(x0)
U1_aag(x0)

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

(26) MRRProof (EQUIVALENT transformation)

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.
Strictly oriented dependency pairs:

U3_GA(ap1_out_aag(X1s, cons(X2s))) → U4_GA(ap2_in_gga(X1s, X2s))
U4_GA(ap2_out_gga(Zs)) → PERM_IN_GA(Zs)
PERM_IN_GA(Xs) → U3_GA(ap1_in_aag(Xs))

Strictly oriented rules of the TRS R:

ap2_in_gga(nil, X) → ap2_out_gga(X)
ap2_in_gga(cons(X), Y) → U2_gga(ap2_in_gga(X, Y))
ap1_in_aag(X) → ap1_out_aag(nil, X)
ap1_in_aag(cons(Z)) → U1_aag(ap1_in_aag(Z))
U2_gga(ap2_out_gga(Z)) → ap2_out_gga(cons(Z))
U1_aag(ap1_out_aag(X, Y)) → ap1_out_aag(cons(X), Y)

Used ordering: Knuth-Bendix order [KBO] with precedence:
ap2ingga2 > nil > PERMINGA1 > cons1 > U4GA1 > ap1inaag1 > U3GA1 > U1aag1 > ap1outaag2 > U2gga1 > ap2outgga1

and weight map:

nil=2
ap2_out_gga_1=6
cons_1=7
U2_gga_1=7
ap1_in_aag_1=2
U1_aag_1=7
U3_GA_1=3
U4_GA_1=1
PERM_IN_GA_1=6
ap2_in_gga_2=8
ap1_out_aag_2=0

The variable weight is 1

(27) Obligation:

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

ap2_in_gga(x0, x1)
ap1_in_aag(x0)
U2_gga(x0)
U1_aag(x0)

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

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) YES