(0) Obligation:
Clauses:
max(X, Y, X) :- less(Y, X).
max(X, Y, Y) :- less(X, s(Y)).
less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).
Queries:
max(a,a,g).
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
max_in: (f,f,b)
less_in: (f,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x2,
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2,
x3)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x2,
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2,
x3)
U2_aag(
x1,
x2,
x3) =
U2_aag(
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:
MAX_IN_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x2,
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2,
x3)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
MAX_IN_AAG(
x1,
x2,
x3) =
MAX_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3) =
U1_AAG(
x1,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
U3_AG(
x1,
x2,
x3) =
U3_AG(
x2,
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x2,
x3)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MAX_IN_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x2,
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2,
x3)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
MAX_IN_AAG(
x1,
x2,
x3) =
MAX_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3) =
U1_AAG(
x1,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
U3_AG(
x1,
x2,
x3) =
U3_AG(
x2,
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x2,
x3)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1,
x2)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x2,
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2,
x3)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
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:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
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:
LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
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:
- LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
The graph contains the following edges 1 > 1
(12) TRUE
(13) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
max_in: (f,f,b)
less_in: (f,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(14) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
(15) 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:
MAX_IN_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
MAX_IN_AAG(
x1,
x2,
x3) =
MAX_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3) =
U1_AAG(
x1,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
U3_AG(
x1,
x2,
x3) =
U3_AG(
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x2,
x3)
We have to consider all (P,R,Pi)-chains
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MAX_IN_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
MAX_IN_AAG(
x1,
x2,
x3) =
MAX_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3) =
U1_AAG(
x1,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
U3_AG(
x1,
x2,
x3) =
U3_AG(
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x2,
x3)
We have to consider all (P,R,Pi)-chains
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
The TRS R consists of the following rules:
max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X1)) → less_out_ag(0, s(X1))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)
The argument filtering Pi contains the following mapping:
max_in_aag(
x1,
x2,
x3) =
max_in_aag(
x3)
U1_aag(
x1,
x2,
x3) =
U1_aag(
x1,
x3)
less_in_ag(
x1,
x2) =
less_in_ag(
x2)
s(
x1) =
s(
x1)
less_out_ag(
x1,
x2) =
less_out_ag(
x1)
U3_ag(
x1,
x2,
x3) =
U3_ag(
x3)
max_out_aag(
x1,
x2,
x3) =
max_out_aag(
x1,
x2)
U2_aag(
x1,
x2,
x3) =
U2_aag(
x2,
x3)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(19) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
LESS_IN_AG(
x1,
x2) =
LESS_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(21) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.