(0) Obligation:
Clauses:sum(X, 0, X).
sum(X, s(Y), s(Z)) :- sum(X, Y, Z).
Queries:
sum(a,g,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:sum1(T5, 0, T5).
sum1(T19, s(0), s(T19)).
sum1(T29, s(s(T27)), s(s(T30))) :- sum1(T29, T27, T30).
Queries:
sum1(a,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:sum1_in: (f,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
sum1_in_aga(T5, 0, T5) → sum1_out_aga(T5, 0, T5)
sum1_in_aga(T19, s(0), s(T19)) → sum1_out_aga(T19, s(0), s(T19))
sum1_in_aga(T29, s(s(T27)), s(s(T30))) → U1_aga(T29, T27, T30, sum1_in_aga(T29, T27, T30))
U1_aga(T29, T27, T30, sum1_out_aga(T29, T27, T30)) → sum1_out_aga(T29, s(s(T27)), s(s(T30)))
The argument filtering Pi contains the following mapping:
sum1_in_aga(x1, x2, x3) = sum1_in_aga(x2)
0 = 0
sum1_out_aga(x1, x2, x3) = sum1_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:The TRS R consists of the following rules:
sum1_in_aga(T5, 0, T5) → sum1_out_aga(T5, 0, T5)
sum1_in_aga(T19, s(0), s(T19)) → sum1_out_aga(T19, s(0), s(T19))
sum1_in_aga(T29, s(s(T27)), s(s(T30))) → U1_aga(T29, T27, T30, sum1_in_aga(T29, T27, T30))
U1_aga(T29, T27, T30, sum1_out_aga(T29, T27, T30)) → sum1_out_aga(T29, s(s(T27)), s(s(T30)))
The argument filtering Pi contains the following mapping:
sum1_in_aga(x1, x2, x3) = sum1_in_aga(x2)
0 = 0
sum1_out_aga(x1, x2, x3) = sum1_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
(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:
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → U1_AGA(T29, T27, T30, sum1_in_aga(T29, T27, T30))
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → SUM1_IN_AGA(T29, T27, T30)
The TRS R consists of the following rules:
sum1_in_aga(T5, 0, T5) → sum1_out_aga(T5, 0, T5)
sum1_in_aga(T19, s(0), s(T19)) → sum1_out_aga(T19, s(0), s(T19))
sum1_in_aga(T29, s(s(T27)), s(s(T30))) → U1_aga(T29, T27, T30, sum1_in_aga(T29, T27, T30))
U1_aga(T29, T27, T30, sum1_out_aga(T29, T27, T30)) → sum1_out_aga(T29, s(s(T27)), s(s(T30)))
The argument filtering Pi contains the following mapping:
sum1_in_aga(x1, x2, x3) = sum1_in_aga(x2)
0 = 0
sum1_out_aga(x1, x2, x3) = sum1_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
SUM1_IN_AGA(x1, x2, x3) = SUM1_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4) = U1_AGA(x4)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → U1_AGA(T29, T27, T30, sum1_in_aga(T29, T27, T30))
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → SUM1_IN_AGA(T29, T27, T30)
The TRS R consists of the following rules:
sum1_in_aga(T5, 0, T5) → sum1_out_aga(T5, 0, T5)
sum1_in_aga(T19, s(0), s(T19)) → sum1_out_aga(T19, s(0), s(T19))
sum1_in_aga(T29, s(s(T27)), s(s(T30))) → U1_aga(T29, T27, T30, sum1_in_aga(T29, T27, T30))
U1_aga(T29, T27, T30, sum1_out_aga(T29, T27, T30)) → sum1_out_aga(T29, s(s(T27)), s(s(T30)))
The argument filtering Pi contains the following mapping:
sum1_in_aga(x1, x2, x3) = sum1_in_aga(x2)
0 = 0
sum1_out_aga(x1, x2, x3) = sum1_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
SUM1_IN_AGA(x1, x2, x3) = SUM1_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4) = U1_AGA(x4)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.(8) Obligation:
Pi DP problem:The TRS P consists of the following rules:
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → SUM1_IN_AGA(T29, T27, T30)
The TRS R consists of the following rules:
sum1_in_aga(T5, 0, T5) → sum1_out_aga(T5, 0, T5)
sum1_in_aga(T19, s(0), s(T19)) → sum1_out_aga(T19, s(0), s(T19))
sum1_in_aga(T29, s(s(T27)), s(s(T30))) → U1_aga(T29, T27, T30, sum1_in_aga(T29, T27, T30))
U1_aga(T29, T27, T30, sum1_out_aga(T29, T27, T30)) → sum1_out_aga(T29, s(s(T27)), s(s(T30)))
The argument filtering Pi contains the following mapping:
sum1_in_aga(x1, x2, x3) = sum1_in_aga(x2)
0 = 0
sum1_out_aga(x1, x2, x3) = sum1_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
SUM1_IN_AGA(x1, x2, x3) = SUM1_IN_AGA(x2)
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:
SUM1_IN_AGA(T29, s(s(T27)), s(s(T30))) → SUM1_IN_AGA(T29, T27, T30)
R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
SUM1_IN_AGA(x1, x2, x3) = SUM1_IN_AGA(x2)
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:
SUM1_IN_AGA(s(s(T27))) → SUM1_IN_AGA(T27)
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:
- SUM1_IN_AGA(s(s(T27))) → SUM1_IN_AGA(T27)
The graph contains the following edges 1 > 1