(0) Obligation:
Clauses:plus(0, Y, Y).
plus(s(X), Y, Z) :- plus(X, s(Y), Z).
Queries:
plus(g,a,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:plus1(0, T5, T5).
plus1(s(0), T18, s(T18)).
plus1(s(s(T25)), T29, T28) :- plus1(T25, s(s(T29)), T28).
Queries:
plus1(g,a,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:plus1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T18, s(T18)) → plus1_out_gaa(s(0), T18, s(T18))
plus1_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plus1_out_gaa(T25, s(s(T29)), T28)) → plus1_out_gaa(s(s(T25)), T29, T28)
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3) = plus1_in_gaa(x1)
0 = 0
plus1_out_gaa(x1, x2, x3) = plus1_out_gaa(x1)
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x1, 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:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T18, s(T18)) → plus1_out_gaa(s(0), T18, s(T18))
plus1_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plus1_out_gaa(T25, s(s(T29)), T28)) → plus1_out_gaa(s(s(T25)), T29, T28)
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3) = plus1_in_gaa(x1)
0 = 0
plus1_out_gaa(x1, x2, x3) = plus1_out_gaa(x1)
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x1, 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:
PLUS1_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T18, s(T18)) → plus1_out_gaa(s(0), T18, s(T18))
plus1_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plus1_out_gaa(T25, s(s(T29)), T28)) → plus1_out_gaa(s(s(T25)), T29, T28)
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3) = plus1_in_gaa(x1)
0 = 0
plus1_out_gaa(x1, x2, x3) = plus1_out_gaa(x1)
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x1, x4)
PLUS1_IN_GAA(x1, x2, x3) = PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x1, x4)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
PLUS1_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T18, s(T18)) → plus1_out_gaa(s(0), T18, s(T18))
plus1_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plus1_out_gaa(T25, s(s(T29)), T28)) → plus1_out_gaa(s(s(T25)), T29, T28)
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3) = plus1_in_gaa(x1)
0 = 0
plus1_out_gaa(x1, x2, x3) = plus1_out_gaa(x1)
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x1, x4)
PLUS1_IN_GAA(x1, x2, x3) = PLUS1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4) = U1_GAA(x1, 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:
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T18, s(T18)) → plus1_out_gaa(s(0), T18, s(T18))
plus1_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plus1_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plus1_out_gaa(T25, s(s(T29)), T28)) → plus1_out_gaa(s(s(T25)), T29, T28)
The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3) = plus1_in_gaa(x1)
0 = 0
plus1_out_gaa(x1, x2, x3) = plus1_out_gaa(x1)
s(x1) = s(x1)
U1_gaa(x1, x2, x3, x4) = U1_gaa(x1, x4)
PLUS1_IN_GAA(x1, x2, x3) = PLUS1_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:
PLUS1_IN_GAA(s(s(T25)), T29, T28) → PLUS1_IN_GAA(T25, s(s(T29)), T28)
R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
PLUS1_IN_GAA(x1, x2, x3) = PLUS1_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:
PLUS1_IN_GAA(s(s(T25))) → PLUS1_IN_GAA(T25)
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:
- PLUS1_IN_GAA(s(s(T25))) → PLUS1_IN_GAA(T25)
The graph contains the following edges 1 > 1