(0) Obligation:

Clauses:

prefix(Xs, Ys) :- app(Xs, X1, Ys).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

prefix(g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

app3([], T12, T12).
app3(.(T19, T20), X32, .(T19, T22)) :- app3(T20, X32, T22).
prefix1(T5, T7) :- app3(T5, X6, T7).

Queries:

prefix1(g,a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
prefix1_in: (b,f)
app3_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

prefix1_in_ga(T5, T7) → U2_ga(T5, T7, app3_in_gaa(T5, X6, T7))
app3_in_gaa([], T12, T12) → app3_out_gaa([], T12, T12)
app3_in_gaa(.(T19, T20), X32, .(T19, T22)) → U1_gaa(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
U1_gaa(T19, T20, X32, T22, app3_out_gaa(T20, X32, T22)) → app3_out_gaa(.(T19, T20), X32, .(T19, T22))
U2_ga(T5, T7, app3_out_gaa(T5, X6, T7)) → prefix1_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefix1_in_ga(x1, x2)  =  prefix1_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app3_in_gaa(x1, x2, x3)  =  app3_in_gaa(x1)
[]  =  []
app3_out_gaa(x1, x2, x3)  =  app3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
prefix1_out_ga(x1, x2)  =  prefix1_out_ga

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

prefix1_in_ga(T5, T7) → U2_ga(T5, T7, app3_in_gaa(T5, X6, T7))
app3_in_gaa([], T12, T12) → app3_out_gaa([], T12, T12)
app3_in_gaa(.(T19, T20), X32, .(T19, T22)) → U1_gaa(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
U1_gaa(T19, T20, X32, T22, app3_out_gaa(T20, X32, T22)) → app3_out_gaa(.(T19, T20), X32, .(T19, T22))
U2_ga(T5, T7, app3_out_gaa(T5, X6, T7)) → prefix1_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefix1_in_ga(x1, x2)  =  prefix1_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app3_in_gaa(x1, x2, x3)  =  app3_in_gaa(x1)
[]  =  []
app3_out_gaa(x1, x2, x3)  =  app3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
prefix1_out_ga(x1, x2)  =  prefix1_out_ga

(5) 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:

PREFIX1_IN_GA(T5, T7) → U2_GA(T5, T7, app3_in_gaa(T5, X6, T7))
PREFIX1_IN_GA(T5, T7) → APP3_IN_GAA(T5, X6, T7)
APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → U1_GAA(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → APP3_IN_GAA(T20, X32, T22)

The TRS R consists of the following rules:

prefix1_in_ga(T5, T7) → U2_ga(T5, T7, app3_in_gaa(T5, X6, T7))
app3_in_gaa([], T12, T12) → app3_out_gaa([], T12, T12)
app3_in_gaa(.(T19, T20), X32, .(T19, T22)) → U1_gaa(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
U1_gaa(T19, T20, X32, T22, app3_out_gaa(T20, X32, T22)) → app3_out_gaa(.(T19, T20), X32, .(T19, T22))
U2_ga(T5, T7, app3_out_gaa(T5, X6, T7)) → prefix1_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefix1_in_ga(x1, x2)  =  prefix1_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app3_in_gaa(x1, x2, x3)  =  app3_in_gaa(x1)
[]  =  []
app3_out_gaa(x1, x2, x3)  =  app3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
prefix1_out_ga(x1, x2)  =  prefix1_out_ga
PREFIX1_IN_GA(x1, x2)  =  PREFIX1_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
APP3_IN_GAA(x1, x2, x3)  =  APP3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(6) Obligation:

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

PREFIX1_IN_GA(T5, T7) → U2_GA(T5, T7, app3_in_gaa(T5, X6, T7))
PREFIX1_IN_GA(T5, T7) → APP3_IN_GAA(T5, X6, T7)
APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → U1_GAA(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → APP3_IN_GAA(T20, X32, T22)

The TRS R consists of the following rules:

prefix1_in_ga(T5, T7) → U2_ga(T5, T7, app3_in_gaa(T5, X6, T7))
app3_in_gaa([], T12, T12) → app3_out_gaa([], T12, T12)
app3_in_gaa(.(T19, T20), X32, .(T19, T22)) → U1_gaa(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
U1_gaa(T19, T20, X32, T22, app3_out_gaa(T20, X32, T22)) → app3_out_gaa(.(T19, T20), X32, .(T19, T22))
U2_ga(T5, T7, app3_out_gaa(T5, X6, T7)) → prefix1_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefix1_in_ga(x1, x2)  =  prefix1_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app3_in_gaa(x1, x2, x3)  =  app3_in_gaa(x1)
[]  =  []
app3_out_gaa(x1, x2, x3)  =  app3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
prefix1_out_ga(x1, x2)  =  prefix1_out_ga
PREFIX1_IN_GA(x1, x2)  =  PREFIX1_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
APP3_IN_GAA(x1, x2, x3)  =  APP3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(8) Obligation:

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

APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → APP3_IN_GAA(T20, X32, T22)

The TRS R consists of the following rules:

prefix1_in_ga(T5, T7) → U2_ga(T5, T7, app3_in_gaa(T5, X6, T7))
app3_in_gaa([], T12, T12) → app3_out_gaa([], T12, T12)
app3_in_gaa(.(T19, T20), X32, .(T19, T22)) → U1_gaa(T19, T20, X32, T22, app3_in_gaa(T20, X32, T22))
U1_gaa(T19, T20, X32, T22, app3_out_gaa(T20, X32, T22)) → app3_out_gaa(.(T19, T20), X32, .(T19, T22))
U2_ga(T5, T7, app3_out_gaa(T5, X6, T7)) → prefix1_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefix1_in_ga(x1, x2)  =  prefix1_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app3_in_gaa(x1, x2, x3)  =  app3_in_gaa(x1)
[]  =  []
app3_out_gaa(x1, x2, x3)  =  app3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
prefix1_out_ga(x1, x2)  =  prefix1_out_ga
APP3_IN_GAA(x1, x2, x3)  =  APP3_IN_GAA(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

APP3_IN_GAA(.(T19, T20), X32, .(T19, T22)) → APP3_IN_GAA(T20, X32, T22)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APP3_IN_GAA(x1, x2, x3)  =  APP3_IN_GAA(x1)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

APP3_IN_GAA(.(T19, T20)) → APP3_IN_GAA(T20)

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

(13) 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:

  • APP3_IN_GAA(.(T19, T20)) → APP3_IN_GAA(T20)
    The graph contains the following edges 1 > 1

(14) YES