(0) Obligation:
Clauses:
p :- r.
r :- ','(!, q).
r :- q.
q.
q :- r.
Queries:
p().
(1) CutEliminatorProof (SOUND transformation)
Eliminated all cuts by simply ignoring them[PROLOG].
(2) Obligation:
Clauses:
p :- r.
r :- q.
r :- q.
q.
q :- r.
Queries:
p().
(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:
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
(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:
P_IN_ → U1_1(r_in_)
P_IN_ → R_IN_
R_IN_ → U2_1(q_in_)
R_IN_ → Q_IN_
Q_IN_ → U3_1(r_in_)
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_ → U1_1(r_in_)
P_IN_ → R_IN_
R_IN_ → U2_1(q_in_)
R_IN_ → Q_IN_
Q_IN_ → U3_1(r_in_)
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
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:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(11) PiDPToQDPProof (EQUIVALENT 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:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
Q_IN_ evaluates to t =
Q_IN_Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceQ_IN_ →
R_IN_with rule
Q_IN_ →
R_IN_ at position [] and matcher [ ]
R_IN_ →
Q_IN_with rule
R_IN_ →
Q_IN_Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(14) FALSE
(15) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(16) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
(17) 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:
P_IN_ → U1_1(r_in_)
P_IN_ → R_IN_
R_IN_ → U2_1(q_in_)
R_IN_ → Q_IN_
Q_IN_ → U3_1(r_in_)
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
We have to consider all (P,R,Pi)-chains
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_ → U1_1(r_in_)
P_IN_ → R_IN_
R_IN_ → U2_1(q_in_)
R_IN_ → Q_IN_
Q_IN_ → U3_1(r_in_)
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
We have to consider all (P,R,Pi)-chains
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
The TRS R consists of the following rules:
p_in_ → U1_(r_in_)
r_in_ → U2_(q_in_)
q_in_ → q_out_
q_in_ → U3_(r_in_)
U3_(r_out_) → q_out_
U2_(q_out_) → r_out_
U1_(r_out_) → p_out_
Pi is empty.
We have to consider all (P,R,Pi)-chains
(21) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(22) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(23) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R_IN_ → Q_IN_
Q_IN_ → R_IN_
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(25) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
Q_IN_ evaluates to t =
Q_IN_Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceQ_IN_ →
R_IN_with rule
Q_IN_ →
R_IN_ at position [] and matcher [ ]
R_IN_ →
Q_IN_with rule
R_IN_ →
Q_IN_Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(26) FALSE