(0) Obligation:
Clauses:
div(X, Y, Z) :- quot(X, Y, Y, Z).
quot(0, s(Y), s(Z), 0).
quot(s(X), s(Y), Z, U) :- quot(X, Y, Z, U).
quot(X, 0, s(Z), s(U)) :- quot(X, s(Z), s(Z), U).
Queries:
div(g,g,a).
(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:
div_in: (b,b,f)
quot_in: (b,b,b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
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:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
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:
DIV_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
DIV_IN_GGA(X, Y, Z) → QUOT_IN_GGGA(X, Y, Y, Z)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x3)
DIV_IN_GGA(
x1,
x2,
x3) =
DIV_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x4)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
U2_GGGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGGA(
x5)
U3_GGGA(
x1,
x2,
x3,
x4) =
U3_GGGA(
x4)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DIV_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
DIV_IN_GGA(X, Y, Z) → QUOT_IN_GGGA(X, Y, Y, Z)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x3)
DIV_IN_GGA(
x1,
x2,
x3) =
DIV_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x4)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
U2_GGGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGGA(
x5)
U3_GGGA(
x1,
x2,
x3,
x4) =
U3_GGGA(
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 4 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x3)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
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:
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
R is empty.
The argument filtering Pi contains the following mapping:
0 =
0
s(
x1) =
s(
x1)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
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:
QUOT_IN_GGGA(X, 0, s(Z)) → QUOT_IN_GGGA(X, s(Z), s(Z))
QUOT_IN_GGGA(s(X), s(Y), Z) → QUOT_IN_GGGA(X, Y, Z)
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:
- QUOT_IN_GGGA(s(X), s(Y), Z) → QUOT_IN_GGGA(X, Y, Z)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3
- QUOT_IN_GGGA(X, 0, s(Z)) → QUOT_IN_GGGA(X, s(Z), s(Z))
The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3
(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:
div_in: (b,b,f)
quot_in: (b,b,b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x1,
x2,
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x1,
x2,
x3,
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x1,
x2,
x3,
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x1,
x2,
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x1,
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:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x1,
x2,
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x1,
x2,
x3,
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x1,
x2,
x3,
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x1,
x2,
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x1,
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:
DIV_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
DIV_IN_GGA(X, Y, Z) → QUOT_IN_GGGA(X, Y, Y, Z)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x1,
x2,
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x1,
x2,
x3,
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x1,
x2,
x3,
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x1,
x2,
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x1,
x2,
x3)
DIV_IN_GGA(
x1,
x2,
x3) =
DIV_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
U2_GGGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGGA(
x1,
x2,
x3,
x5)
U3_GGGA(
x1,
x2,
x3,
x4) =
U3_GGGA(
x1,
x2,
x4)
We have to consider all (P,R,Pi)-chains
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DIV_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
DIV_IN_GGA(X, Y, Z) → QUOT_IN_GGGA(X, Y, Y, Z)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x1,
x2,
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x1,
x2,
x3,
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x1,
x2,
x3,
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x1,
x2,
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x1,
x2,
x3)
DIV_IN_GGA(
x1,
x2,
x3) =
DIV_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
U2_GGGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGGA(
x1,
x2,
x3,
x5)
U3_GGGA(
x1,
x2,
x3,
x4) =
U3_GGGA(
x1,
x2,
x4)
We have to consider all (P,R,Pi)-chains
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
The TRS R consists of the following rules:
div_in_gga(X, Y, Z) → U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z))
quot_in_ggga(0, s(Y), s(Z), 0) → quot_out_ggga(0, s(Y), s(Z), 0)
quot_in_ggga(s(X), s(Y), Z, U) → U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U))
quot_in_ggga(X, 0, s(Z), s(U)) → U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U))
U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) → quot_out_ggga(X, 0, s(Z), s(U))
U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) → quot_out_ggga(s(X), s(Y), Z, U)
U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) → div_out_gga(X, Y, Z)
The argument filtering Pi contains the following mapping:
div_in_gga(
x1,
x2,
x3) =
div_in_gga(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4) =
U1_gga(
x1,
x2,
x4)
quot_in_ggga(
x1,
x2,
x3,
x4) =
quot_in_ggga(
x1,
x2,
x3)
0 =
0
s(
x1) =
s(
x1)
quot_out_ggga(
x1,
x2,
x3,
x4) =
quot_out_ggga(
x1,
x2,
x3,
x4)
U2_ggga(
x1,
x2,
x3,
x4,
x5) =
U2_ggga(
x1,
x2,
x3,
x5)
U3_ggga(
x1,
x2,
x3,
x4) =
U3_ggga(
x1,
x2,
x4)
div_out_gga(
x1,
x2,
x3) =
div_out_gga(
x1,
x2,
x3)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
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:
QUOT_IN_GGGA(X, 0, s(Z), s(U)) → QUOT_IN_GGGA(X, s(Z), s(Z), U)
QUOT_IN_GGGA(s(X), s(Y), Z, U) → QUOT_IN_GGGA(X, Y, Z, U)
R is empty.
The argument filtering Pi contains the following mapping:
0 =
0
s(
x1) =
s(
x1)
QUOT_IN_GGGA(
x1,
x2,
x3,
x4) =
QUOT_IN_GGGA(
x1,
x2,
x3)
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:
QUOT_IN_GGGA(X, 0, s(Z)) → QUOT_IN_GGGA(X, s(Z), s(Z))
QUOT_IN_GGGA(s(X), s(Y), Z) → QUOT_IN_GGGA(X, Y, Z)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.