(0) Obligation:
Clauses:
sum(X, s(Y), s(Z)) :- sum(X, Y, Z).
sum(X, 0, X).
Query: sum(a,a,g)
(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)
Transformed Prolog program to (Pi-)TRS.
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
sumA_in_aag(T27, s(s(T28)), s(s(T26))) → U1_aag(T27, T28, T26, sumA_in_aag(T27, T28, T26))
sumA_in_aag(T36, s(0), s(T36)) → sumA_out_aag(T36, s(0), s(T36))
sumA_in_aag(s(T42), 0, s(T42)) → sumA_out_aag(s(T42), 0, s(T42))
sumA_in_aag(T44, 0, T44) → sumA_out_aag(T44, 0, T44)
U1_aag(T27, T28, T26, sumA_out_aag(T27, T28, T26)) → sumA_out_aag(T27, s(s(T28)), s(s(T26)))
The argument filtering Pi contains the following mapping:
sumA_in_aag(
x1,
x2,
x3) =
sumA_in_aag(
x3)
s(
x1) =
s(
x1)
U1_aag(
x1,
x2,
x3,
x4) =
U1_aag(
x3,
x4)
sumA_out_aag(
x1,
x2,
x3) =
sumA_out_aag(
x1,
x2,
x3)
(3) 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:
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → U1_AAG(T27, T28, T26, sumA_in_aag(T27, T28, T26))
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → SUMA_IN_AAG(T27, T28, T26)
The TRS R consists of the following rules:
sumA_in_aag(T27, s(s(T28)), s(s(T26))) → U1_aag(T27, T28, T26, sumA_in_aag(T27, T28, T26))
sumA_in_aag(T36, s(0), s(T36)) → sumA_out_aag(T36, s(0), s(T36))
sumA_in_aag(s(T42), 0, s(T42)) → sumA_out_aag(s(T42), 0, s(T42))
sumA_in_aag(T44, 0, T44) → sumA_out_aag(T44, 0, T44)
U1_aag(T27, T28, T26, sumA_out_aag(T27, T28, T26)) → sumA_out_aag(T27, s(s(T28)), s(s(T26)))
The argument filtering Pi contains the following mapping:
sumA_in_aag(
x1,
x2,
x3) =
sumA_in_aag(
x3)
s(
x1) =
s(
x1)
U1_aag(
x1,
x2,
x3,
x4) =
U1_aag(
x3,
x4)
sumA_out_aag(
x1,
x2,
x3) =
sumA_out_aag(
x1,
x2,
x3)
SUMA_IN_AAG(
x1,
x2,
x3) =
SUMA_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3,
x4) =
U1_AAG(
x3,
x4)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → U1_AAG(T27, T28, T26, sumA_in_aag(T27, T28, T26))
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → SUMA_IN_AAG(T27, T28, T26)
The TRS R consists of the following rules:
sumA_in_aag(T27, s(s(T28)), s(s(T26))) → U1_aag(T27, T28, T26, sumA_in_aag(T27, T28, T26))
sumA_in_aag(T36, s(0), s(T36)) → sumA_out_aag(T36, s(0), s(T36))
sumA_in_aag(s(T42), 0, s(T42)) → sumA_out_aag(s(T42), 0, s(T42))
sumA_in_aag(T44, 0, T44) → sumA_out_aag(T44, 0, T44)
U1_aag(T27, T28, T26, sumA_out_aag(T27, T28, T26)) → sumA_out_aag(T27, s(s(T28)), s(s(T26)))
The argument filtering Pi contains the following mapping:
sumA_in_aag(
x1,
x2,
x3) =
sumA_in_aag(
x3)
s(
x1) =
s(
x1)
U1_aag(
x1,
x2,
x3,
x4) =
U1_aag(
x3,
x4)
sumA_out_aag(
x1,
x2,
x3) =
sumA_out_aag(
x1,
x2,
x3)
SUMA_IN_AAG(
x1,
x2,
x3) =
SUMA_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3,
x4) =
U1_AAG(
x3,
x4)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → SUMA_IN_AAG(T27, T28, T26)
The TRS R consists of the following rules:
sumA_in_aag(T27, s(s(T28)), s(s(T26))) → U1_aag(T27, T28, T26, sumA_in_aag(T27, T28, T26))
sumA_in_aag(T36, s(0), s(T36)) → sumA_out_aag(T36, s(0), s(T36))
sumA_in_aag(s(T42), 0, s(T42)) → sumA_out_aag(s(T42), 0, s(T42))
sumA_in_aag(T44, 0, T44) → sumA_out_aag(T44, 0, T44)
U1_aag(T27, T28, T26, sumA_out_aag(T27, T28, T26)) → sumA_out_aag(T27, s(s(T28)), s(s(T26)))
The argument filtering Pi contains the following mapping:
sumA_in_aag(
x1,
x2,
x3) =
sumA_in_aag(
x3)
s(
x1) =
s(
x1)
U1_aag(
x1,
x2,
x3,
x4) =
U1_aag(
x3,
x4)
sumA_out_aag(
x1,
x2,
x3) =
sumA_out_aag(
x1,
x2,
x3)
SUMA_IN_AAG(
x1,
x2,
x3) =
SUMA_IN_AAG(
x3)
We have to consider all (P,R,Pi)-chains
(7) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
SUMA_IN_AAG(T27, s(s(T28)), s(s(T26))) → SUMA_IN_AAG(T27, T28, T26)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
SUMA_IN_AAG(
x1,
x2,
x3) =
SUMA_IN_AAG(
x3)
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SUMA_IN_AAG(s(s(T26))) → SUMA_IN_AAG(T26)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) 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:
- SUMA_IN_AAG(s(s(T26))) → SUMA_IN_AAG(T26)
The graph contains the following edges 1 > 1
(12) YES