(0) Obligation:

Clauses:

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

Queries:

suffix(a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

app3([], T12, T12).
app3(.(X29, X30), T19, .(X29, T18)) :- app3(X30, T19, T18).
suffix1(T7, T6) :- app3(X6, T7, T6).

Queries:

suffix1(a,g).

(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:
suffix1_in: (f,b)
app3_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

suffix1_in_ag(T7, T6) → U2_ag(T7, T6, app3_in_aag(X6, T7, T6))
app3_in_aag([], T12, T12) → app3_out_aag([], T12, T12)
app3_in_aag(.(X29, X30), T19, .(X29, T18)) → U1_aag(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
U1_aag(X29, X30, T19, T18, app3_out_aag(X30, T19, T18)) → app3_out_aag(.(X29, X30), T19, .(X29, T18))
U2_ag(T7, T6, app3_out_aag(X6, T7, T6)) → suffix1_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffix1_in_ag(x1, x2)  =  suffix1_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
app3_in_aag(x1, x2, x3)  =  app3_in_aag(x3)
app3_out_aag(x1, x2, x3)  =  app3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffix1_out_ag(x1, x2)  =  suffix1_out_ag(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

suffix1_in_ag(T7, T6) → U2_ag(T7, T6, app3_in_aag(X6, T7, T6))
app3_in_aag([], T12, T12) → app3_out_aag([], T12, T12)
app3_in_aag(.(X29, X30), T19, .(X29, T18)) → U1_aag(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
U1_aag(X29, X30, T19, T18, app3_out_aag(X30, T19, T18)) → app3_out_aag(.(X29, X30), T19, .(X29, T18))
U2_ag(T7, T6, app3_out_aag(X6, T7, T6)) → suffix1_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffix1_in_ag(x1, x2)  =  suffix1_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
app3_in_aag(x1, x2, x3)  =  app3_in_aag(x3)
app3_out_aag(x1, x2, x3)  =  app3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffix1_out_ag(x1, x2)  =  suffix1_out_ag(x1)

(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:

SUFFIX1_IN_AG(T7, T6) → U2_AG(T7, T6, app3_in_aag(X6, T7, T6))
SUFFIX1_IN_AG(T7, T6) → APP3_IN_AAG(X6, T7, T6)
APP3_IN_AAG(.(X29, X30), T19, .(X29, T18)) → U1_AAG(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
APP3_IN_AAG(.(X29, X30), T19, .(X29, T18)) → APP3_IN_AAG(X30, T19, T18)

The TRS R consists of the following rules:

suffix1_in_ag(T7, T6) → U2_ag(T7, T6, app3_in_aag(X6, T7, T6))
app3_in_aag([], T12, T12) → app3_out_aag([], T12, T12)
app3_in_aag(.(X29, X30), T19, .(X29, T18)) → U1_aag(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
U1_aag(X29, X30, T19, T18, app3_out_aag(X30, T19, T18)) → app3_out_aag(.(X29, X30), T19, .(X29, T18))
U2_ag(T7, T6, app3_out_aag(X6, T7, T6)) → suffix1_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffix1_in_ag(x1, x2)  =  suffix1_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
app3_in_aag(x1, x2, x3)  =  app3_in_aag(x3)
app3_out_aag(x1, x2, x3)  =  app3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffix1_out_ag(x1, x2)  =  suffix1_out_ag(x1)
SUFFIX1_IN_AG(x1, x2)  =  SUFFIX1_IN_AG(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
APP3_IN_AAG(x1, x2, x3)  =  APP3_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)

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

(6) Obligation:

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

SUFFIX1_IN_AG(T7, T6) → U2_AG(T7, T6, app3_in_aag(X6, T7, T6))
SUFFIX1_IN_AG(T7, T6) → APP3_IN_AAG(X6, T7, T6)
APP3_IN_AAG(.(X29, X30), T19, .(X29, T18)) → U1_AAG(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
APP3_IN_AAG(.(X29, X30), T19, .(X29, T18)) → APP3_IN_AAG(X30, T19, T18)

The TRS R consists of the following rules:

suffix1_in_ag(T7, T6) → U2_ag(T7, T6, app3_in_aag(X6, T7, T6))
app3_in_aag([], T12, T12) → app3_out_aag([], T12, T12)
app3_in_aag(.(X29, X30), T19, .(X29, T18)) → U1_aag(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
U1_aag(X29, X30, T19, T18, app3_out_aag(X30, T19, T18)) → app3_out_aag(.(X29, X30), T19, .(X29, T18))
U2_ag(T7, T6, app3_out_aag(X6, T7, T6)) → suffix1_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffix1_in_ag(x1, x2)  =  suffix1_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
app3_in_aag(x1, x2, x3)  =  app3_in_aag(x3)
app3_out_aag(x1, x2, x3)  =  app3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffix1_out_ag(x1, x2)  =  suffix1_out_ag(x1)
SUFFIX1_IN_AG(x1, x2)  =  SUFFIX1_IN_AG(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
APP3_IN_AAG(x1, x2, x3)  =  APP3_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, 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_AAG(.(X29, X30), T19, .(X29, T18)) → APP3_IN_AAG(X30, T19, T18)

The TRS R consists of the following rules:

suffix1_in_ag(T7, T6) → U2_ag(T7, T6, app3_in_aag(X6, T7, T6))
app3_in_aag([], T12, T12) → app3_out_aag([], T12, T12)
app3_in_aag(.(X29, X30), T19, .(X29, T18)) → U1_aag(X29, X30, T19, T18, app3_in_aag(X30, T19, T18))
U1_aag(X29, X30, T19, T18, app3_out_aag(X30, T19, T18)) → app3_out_aag(.(X29, X30), T19, .(X29, T18))
U2_ag(T7, T6, app3_out_aag(X6, T7, T6)) → suffix1_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffix1_in_ag(x1, x2)  =  suffix1_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
app3_in_aag(x1, x2, x3)  =  app3_in_aag(x3)
app3_out_aag(x1, x2, x3)  =  app3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffix1_out_ag(x1, x2)  =  suffix1_out_ag(x1)
APP3_IN_AAG(x1, x2, x3)  =  APP3_IN_AAG(x3)

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_AAG(.(X29, X30), T19, .(X29, T18)) → APP3_IN_AAG(X30, T19, T18)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APP3_IN_AAG(x1, x2, x3)  =  APP3_IN_AAG(x3)

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_AAG(.(X29, T18)) → APP3_IN_AAG(T18)

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_AAG(.(X29, T18)) → APP3_IN_AAG(T18)
    The graph contains the following edges 1 > 1

(14) YES