(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