(0) Obligation:
Clauses:color_map(.(Region, Regions), Colors) :- ','(color_region(Region, Colors), color_map(Regions, Colors)).
color_map([], Colors).
color_region(region(Color, Neighbors), Colors) :- ','(select(Color, Colors, Colors1), members(Neighbors, Colors1)).
select(X, .(X, Xs), Xs).
select(X, .(Y, Ys), .(Y, Zs)) :- select(X, Ys, Zs).
members(.(X, Xs), Ys) :- ','(member(X, Ys), members(Xs, Ys)).
members([], Ys).
member(X, .(X, X1)).
member(X, .(X2, T)) :- member(X, T).
test_color(Name, Pairs) :- ','(colors(Name, Colors), ','(color_map(Map, Colors), ','(map(Name, Symbols, Map), symbols(Symbols, Map, Pairs)))).
symbols([], [], []).
symbols(.(S, Ss), .(region(C, N), Rs), .(pair(S, C), Ps)) :- symbols(Ss, Rs, Ps).
map(test, .(a, .(b, .(c, .(d, .(e, .(f, [])))))), .(region(A, .(B, .(C, .(D, [])))), .(region(B, .(A, .(C, .(E, [])))), .(region(C, .(A, .(B, .(D, .(E, .(F, [])))))), .(region(D, .(A, .(C, .(F, [])))), .(region(E, .(B, .(C, .(F, [])))), .(region(F, .(C, .(D, .(E, [])))), []))))))).
map(west_europe, .(portugal, .(spain, .(france, .(belgium, .(holland, .(west_germany, .(luxembourg, .(italy, .(switzerland, .(austria, [])))))))))), .(region(P, .(E, [])), .(region(E, .(F, .(P, []))), .(region(F, .(E, .(I, .(S, .(B, .(WG, .(L, []))))))), .(region(B, .(F, .(H, .(L, .(WG, []))))), .(region(H, .(B, .(WG, []))), .(region(WG, .(F, .(A, .(S, .(H, .(B, .(L, []))))))), .(region(L, .(F, .(B, .(WG, [])))), .(region(I, .(F, .(A, .(S, [])))), .(region(S, .(F, .(I, .(A, .(WG, []))))), .(region(A, .(I, .(S, .(WG, [])))), []))))))))))).
colors(X, .(red, .(yellow, .(blue, .(white, []))))).
Queries:
color_map(a,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:select10(T59, .(T57, T58), .(T57, X67)) :- select10(T59, T58, X67).
members20(.(T86, T87), T85) :- member27(T86, T85).
members20(.(T86, T92), T85) :- ','(memberc27(T86, T85), members20(T92, T85)).
member27(T116, .(T114, T115)) :- member27(T116, T115).
color_map1(.(region(T26, T27), T28), T25) :- select10(T26, T25, X29).
color_map1(.(region(T26, T34), T35), T25) :- ','(selectc10(T26, T25, T33), members20(T34, T33)).
color_map1(.(region(T26, T34), T67), T25) :- ','(selectc10(T26, T25, T33), ','(membersc20(T34, T33), color_map1(T67, T25))).
Clauses:
selectc10(T48, .(T48, T49), T49).
selectc10(T59, .(T57, T58), .(T57, X67)) :- selectc10(T59, T58, X67).
color_mapc1(.(region(T26, T34), T67), T25) :- ','(selectc10(T26, T25, T33), ','(membersc20(T34, T33), color_mapc1(T67, T25))).
color_mapc1([], T132).
color_mapc1([], T134).
membersc20(.(T86, T92), T85) :- ','(memberc27(T86, T85), membersc20(T92, T85)).
membersc20([], T126).
memberc27(T105, .(T105, T106)).
memberc27(T116, .(T114, T115)) :- memberc27(T116, T115).
Afs:
color_map1(x1, x2) = color_map1(x2)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:color_map1_in: (f,b)
select10_in: (f,b,f)
selectc10_in: (f,b,f)
members20_in: (f,b)
member27_in: (f,b)
memberc27_in: (f,b)
membersc20_in: (f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(region(T26, T27), T28), T25) → U6_AG(T26, T27, T28, T25, select10_in_aga(T26, T25, X29))
COLOR_MAP1_IN_AG(.(region(T26, T27), T28), T25) → SELECT10_IN_AGA(T26, T25, X29)
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → U1_AGA(T59, T57, T58, X67, select10_in_aga(T59, T58, X67))
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → SELECT10_IN_AGA(T59, T58, X67)
COLOR_MAP1_IN_AG(.(region(T26, T34), T35), T25) → U7_AG(T26, T34, T35, T25, selectc10_in_aga(T26, T25, T33))
U7_AG(T26, T34, T35, T25, selectc10_out_aga(T26, T25, T33)) → U8_AG(T26, T34, T35, T25, members20_in_ag(T34, T33))
U7_AG(T26, T34, T35, T25, selectc10_out_aga(T26, T25, T33)) → MEMBERS20_IN_AG(T34, T33)
MEMBERS20_IN_AG(.(T86, T87), T85) → U2_AG(T86, T87, T85, member27_in_ag(T86, T85))
MEMBERS20_IN_AG(.(T86, T87), T85) → MEMBER27_IN_AG(T86, T85)
MEMBER27_IN_AG(T116, .(T114, T115)) → U5_AG(T116, T114, T115, member27_in_ag(T116, T115))
MEMBER27_IN_AG(T116, .(T114, T115)) → MEMBER27_IN_AG(T116, T115)
MEMBERS20_IN_AG(.(T86, T92), T85) → U3_AG(T86, T92, T85, memberc27_in_ag(T86, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → U4_AG(T86, T92, T85, members20_in_ag(T92, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T92, T85)
COLOR_MAP1_IN_AG(.(region(T26, T34), T67), T25) → U9_AG(T26, T34, T67, T25, selectc10_in_aga(T26, T25, T33))
U9_AG(T26, T34, T67, T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T26, T34, T67, T25, membersc20_in_ag(T34, T33))
U10_AG(T26, T34, T67, T25, membersc20_out_ag(T34, T33)) → U11_AG(T26, T34, T67, T25, color_map1_in_ag(T67, T25))
U10_AG(T26, T34, T67, T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T67, T25)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
color_map1_in_ag(x1, x2) = color_map1_in_ag(x2)
select10_in_aga(x1, x2, x3) = select10_in_aga(x2)
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
members20_in_ag(x1, x2) = members20_in_ag(x2)
member27_in_ag(x1, x2) = member27_in_ag(x2)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
COLOR_MAP1_IN_AG(x1, x2) = COLOR_MAP1_IN_AG(x2)
U6_AG(x1, x2, x3, x4, x5) = U6_AG(x4, x5)
SELECT10_IN_AGA(x1, x2, x3) = SELECT10_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5) = U1_AGA(x2, x3, x5)
U7_AG(x1, x2, x3, x4, x5) = U7_AG(x4, x5)
U8_AG(x1, x2, x3, x4, x5) = U8_AG(x4, x5)
MEMBERS20_IN_AG(x1, x2) = MEMBERS20_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
MEMBER27_IN_AG(x1, x2) = MEMBER27_IN_AG(x2)
U5_AG(x1, x2, x3, x4) = U5_AG(x2, x3, x4)
U3_AG(x1, x2, x3, x4) = U3_AG(x3, x4)
U4_AG(x1, x2, x3, x4) = U4_AG(x3, x4)
U9_AG(x1, x2, x3, x4, x5) = U9_AG(x4, x5)
U10_AG(x1, x2, x3, x4, x5) = U10_AG(x4, x5)
U11_AG(x1, x2, x3, x4, x5) = U11_AG(x4, x5)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(region(T26, T27), T28), T25) → U6_AG(T26, T27, T28, T25, select10_in_aga(T26, T25, X29))
COLOR_MAP1_IN_AG(.(region(T26, T27), T28), T25) → SELECT10_IN_AGA(T26, T25, X29)
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → U1_AGA(T59, T57, T58, X67, select10_in_aga(T59, T58, X67))
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → SELECT10_IN_AGA(T59, T58, X67)
COLOR_MAP1_IN_AG(.(region(T26, T34), T35), T25) → U7_AG(T26, T34, T35, T25, selectc10_in_aga(T26, T25, T33))
U7_AG(T26, T34, T35, T25, selectc10_out_aga(T26, T25, T33)) → U8_AG(T26, T34, T35, T25, members20_in_ag(T34, T33))
U7_AG(T26, T34, T35, T25, selectc10_out_aga(T26, T25, T33)) → MEMBERS20_IN_AG(T34, T33)
MEMBERS20_IN_AG(.(T86, T87), T85) → U2_AG(T86, T87, T85, member27_in_ag(T86, T85))
MEMBERS20_IN_AG(.(T86, T87), T85) → MEMBER27_IN_AG(T86, T85)
MEMBER27_IN_AG(T116, .(T114, T115)) → U5_AG(T116, T114, T115, member27_in_ag(T116, T115))
MEMBER27_IN_AG(T116, .(T114, T115)) → MEMBER27_IN_AG(T116, T115)
MEMBERS20_IN_AG(.(T86, T92), T85) → U3_AG(T86, T92, T85, memberc27_in_ag(T86, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → U4_AG(T86, T92, T85, members20_in_ag(T92, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T92, T85)
COLOR_MAP1_IN_AG(.(region(T26, T34), T67), T25) → U9_AG(T26, T34, T67, T25, selectc10_in_aga(T26, T25, T33))
U9_AG(T26, T34, T67, T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T26, T34, T67, T25, membersc20_in_ag(T34, T33))
U10_AG(T26, T34, T67, T25, membersc20_out_ag(T34, T33)) → U11_AG(T26, T34, T67, T25, color_map1_in_ag(T67, T25))
U10_AG(T26, T34, T67, T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T67, T25)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
color_map1_in_ag(x1, x2) = color_map1_in_ag(x2)
select10_in_aga(x1, x2, x3) = select10_in_aga(x2)
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
members20_in_ag(x1, x2) = members20_in_ag(x2)
member27_in_ag(x1, x2) = member27_in_ag(x2)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
COLOR_MAP1_IN_AG(x1, x2) = COLOR_MAP1_IN_AG(x2)
U6_AG(x1, x2, x3, x4, x5) = U6_AG(x4, x5)
SELECT10_IN_AGA(x1, x2, x3) = SELECT10_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5) = U1_AGA(x2, x3, x5)
U7_AG(x1, x2, x3, x4, x5) = U7_AG(x4, x5)
U8_AG(x1, x2, x3, x4, x5) = U8_AG(x4, x5)
MEMBERS20_IN_AG(x1, x2) = MEMBERS20_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
MEMBER27_IN_AG(x1, x2) = MEMBER27_IN_AG(x2)
U5_AG(x1, x2, x3, x4) = U5_AG(x2, x3, x4)
U3_AG(x1, x2, x3, x4) = U3_AG(x3, x4)
U4_AG(x1, x2, x3, x4) = U4_AG(x3, x4)
U9_AG(x1, x2, x3, x4, x5) = U9_AG(x4, x5)
U10_AG(x1, x2, x3, x4, x5) = U10_AG(x4, x5)
U11_AG(x1, x2, x3, x4, x5) = U11_AG(x4, x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 4 SCCs with 11 less nodes.(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER27_IN_AG(T116, .(T114, T115)) → MEMBER27_IN_AG(T116, T115)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
MEMBER27_IN_AG(x1, x2) = MEMBER27_IN_AG(x2)
We have to consider all (P,R,Pi)-chains
(8) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.(9) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER27_IN_AG(T116, .(T114, T115)) → MEMBER27_IN_AG(T116, T115)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
MEMBER27_IN_AG(x1, x2) = MEMBER27_IN_AG(x2)
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(11) Obligation:
Q DP problem:The TRS P consists of the following rules:
MEMBER27_IN_AG(.(T114, T115)) → MEMBER27_IN_AG(T115)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) 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:
- MEMBER27_IN_AG(.(T114, T115)) → MEMBER27_IN_AG(T115)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBERS20_IN_AG(.(T86, T92), T85) → U3_AG(T86, T92, T85, memberc27_in_ag(T86, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T92, T85)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
MEMBERS20_IN_AG(x1, x2) = MEMBERS20_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x3, x4)
We have to consider all (P,R,Pi)-chains
(15) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.(16) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBERS20_IN_AG(.(T86, T92), T85) → U3_AG(T86, T92, T85, memberc27_in_ag(T86, T85))
U3_AG(T86, T92, T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T92, T85)
The TRS R consists of the following rules:
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
MEMBERS20_IN_AG(x1, x2) = MEMBERS20_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x3, x4)
We have to consider all (P,R,Pi)-chains
(17) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(18) Obligation:
Q DP problem:The TRS P consists of the following rules:
MEMBERS20_IN_AG(T85) → U3_AG(T85, memberc27_in_ag(T85))
U3_AG(T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T85)
The TRS R consists of the following rules:
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
The set Q consists of the following terms:
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(19) Narrowing (SOUND transformation)
By narrowing [LPAR04] the rule MEMBERS20_IN_AG(T85) → U3_AG(T85, memberc27_in_ag(T85)) at position [1] we obtained the following new rules [LPAR04]:
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), memberc27_out_ag(x0, .(x0, x1)))
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), U19_ag(x0, x1, memberc27_in_ag(x1)))
(20) Obligation:
Q DP problem:The TRS P consists of the following rules:
U3_AG(T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T85)
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), memberc27_out_ag(x0, .(x0, x1)))
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), U19_ag(x0, x1, memberc27_in_ag(x1)))
The TRS R consists of the following rules:
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
The set Q consists of the following terms:
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(21) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule U3_AG(T85, memberc27_out_ag(T86, T85)) → MEMBERS20_IN_AG(T85) we obtained the following new rules [LPAR04]:
U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1))) → MEMBERS20_IN_AG(.(z0, z1))
U3_AG(.(z0, z1), memberc27_out_ag(x1, .(z0, z1))) → MEMBERS20_IN_AG(.(z0, z1))
(22) Obligation:
Q DP problem:The TRS P consists of the following rules:
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), memberc27_out_ag(x0, .(x0, x1)))
MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), U19_ag(x0, x1, memberc27_in_ag(x1)))
U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1))) → MEMBERS20_IN_AG(.(z0, z1))
U3_AG(.(z0, z1), memberc27_out_ag(x1, .(z0, z1))) → MEMBERS20_IN_AG(.(z0, z1))
The TRS R consists of the following rules:
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
The set Q consists of the following terms:
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(23) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.Found a loop by narrowing to the left:
s = U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1))) evaluates to t =U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1)))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequence
U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1))) → MEMBERS20_IN_AG(.(z0, z1))
with rule U3_AG(.(z0', z1'), memberc27_out_ag(z0', .(z0', z1'))) → MEMBERS20_IN_AG(.(z0', z1')) at position [] and matcher [z0' / z0, z1' / z1]
MEMBERS20_IN_AG(.(z0, z1)) → U3_AG(.(z0, z1), memberc27_out_ag(z0, .(z0, z1)))
with rule MEMBERS20_IN_AG(.(x0, x1)) → U3_AG(.(x0, x1), memberc27_out_ag(x0, .(x0, x1)))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(24) NO
(25) Obligation:
Pi DP problem:The TRS P consists of the following rules:
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → SELECT10_IN_AGA(T59, T58, X67)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
SELECT10_IN_AGA(x1, x2, x3) = SELECT10_IN_AGA(x2)
We have to consider all (P,R,Pi)-chains
(26) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.(27) Obligation:
Pi DP problem:The TRS P consists of the following rules:
SELECT10_IN_AGA(T59, .(T57, T58), .(T57, X67)) → SELECT10_IN_AGA(T59, T58, X67)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
SELECT10_IN_AGA(x1, x2, x3) = SELECT10_IN_AGA(x2)
We have to consider all (P,R,Pi)-chains
(28) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(29) Obligation:
Q DP problem:The TRS P consists of the following rules:
SELECT10_IN_AGA(.(T57, T58)) → SELECT10_IN_AGA(T58)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(30) 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:
- SELECT10_IN_AGA(.(T57, T58)) → SELECT10_IN_AGA(T58)
The graph contains the following edges 1 > 1
(31) YES
(32) Obligation:
Pi DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(region(T26, T34), T67), T25) → U9_AG(T26, T34, T67, T25, selectc10_in_aga(T26, T25, T33))
U9_AG(T26, T34, T67, T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T26, T34, T67, T25, membersc20_in_ag(T34, T33))
U10_AG(T26, T34, T67, T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T67, T25)
The TRS R consists of the following rules:
selectc10_in_aga(T48, .(T48, T49), T49) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(T59, .(T57, T58), .(T57, X67)) → U13_aga(T59, T57, T58, X67, selectc10_in_aga(T59, T58, X67))
U13_aga(T59, T57, T58, X67, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(T105, .(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(T116, .(T114, T115)) → U19_ag(T116, T114, T115, memberc27_in_ag(T116, T115))
U19_ag(T116, T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(.(T86, T92), T85) → U17_ag(T86, T92, T85, memberc27_in_ag(T86, T85))
U17_ag(T86, T92, T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T92, T85, membersc20_in_ag(T92, T85))
membersc20_in_ag([], T126) → membersc20_out_ag([], T126)
U18_ag(T86, T92, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
selectc10_in_aga(x1, x2, x3) = selectc10_in_aga(x2)
selectc10_out_aga(x1, x2, x3) = selectc10_out_aga(x1, x2, x3)
U13_aga(x1, x2, x3, x4, x5) = U13_aga(x2, x3, x5)
memberc27_in_ag(x1, x2) = memberc27_in_ag(x2)
memberc27_out_ag(x1, x2) = memberc27_out_ag(x1, x2)
U19_ag(x1, x2, x3, x4) = U19_ag(x2, x3, x4)
membersc20_in_ag(x1, x2) = membersc20_in_ag(x2)
U17_ag(x1, x2, x3, x4) = U17_ag(x3, x4)
U18_ag(x1, x2, x3, x4) = U18_ag(x1, x3, x4)
membersc20_out_ag(x1, x2) = membersc20_out_ag(x1, x2)
COLOR_MAP1_IN_AG(x1, x2) = COLOR_MAP1_IN_AG(x2)
U9_AG(x1, x2, x3, x4, x5) = U9_AG(x4, x5)
U10_AG(x1, x2, x3, x4, x5) = U10_AG(x4, x5)
We have to consider all (P,R,Pi)-chains
(33) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(34) Obligation:
Q DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(T25) → U9_AG(T25, selectc10_in_aga(T25))
U9_AG(T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T25, membersc20_in_ag(T33))
U10_AG(T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T25)
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(35) Narrowing (SOUND transformation)
By narrowing [LPAR04] the rule COLOR_MAP1_IN_AG(T25) → U9_AG(T25, selectc10_in_aga(T25)) at position [1] we obtained the following new rules [LPAR04]:
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
(36) Obligation:
Q DP problem:The TRS P consists of the following rules:
U9_AG(T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T25, membersc20_in_ag(T33))
U10_AG(T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T25)
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(37) Narrowing (SOUND transformation)
By narrowing [LPAR04] the rule U9_AG(T25, selectc10_out_aga(T26, T25, T33)) → U10_AG(T25, membersc20_in_ag(T33)) at position [1] we obtained the following new rules [LPAR04]:
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, U17_ag(x0, memberc27_in_ag(x0)))
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, membersc20_out_ag([], x0))
(38) Obligation:
Q DP problem:The TRS P consists of the following rules:
U10_AG(T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T25)
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, U17_ag(x0, memberc27_in_ag(x0)))
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, membersc20_out_ag([], x0))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(39) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, U17_ag(x0, memberc27_in_ag(x0))) we obtained the following new rules [LPAR04]:
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), U17_ag(z1, memberc27_in_ag(z1)))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), U17_ag(x2, memberc27_in_ag(x2)))
(40) Obligation:
Q DP problem:The TRS P consists of the following rules:
U10_AG(T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T25)
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, membersc20_out_ag([], x0))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), U17_ag(z1, memberc27_in_ag(z1)))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), U17_ag(x2, memberc27_in_ag(x2)))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(41) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule U10_AG(T25, membersc20_out_ag(T34, T33)) → COLOR_MAP1_IN_AG(T25) we obtained the following new rules [LPAR04]:
U10_AG(z0, membersc20_out_ag([], z2)) → COLOR_MAP1_IN_AG(z0)
U10_AG(.(z0, z1), membersc20_out_ag(x1, x2)) → COLOR_MAP1_IN_AG(.(z0, z1))
(42) Obligation:
Q DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, membersc20_out_ag([], x0))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), U17_ag(z1, memberc27_in_ag(z1)))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), U17_ag(x2, memberc27_in_ag(x2)))
U10_AG(z0, membersc20_out_ag([], z2)) → COLOR_MAP1_IN_AG(z0)
U10_AG(.(z0, z1), membersc20_out_ag(x1, x2)) → COLOR_MAP1_IN_AG(.(z0, z1))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(43) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule U9_AG(y0, selectc10_out_aga(y1, y0, x0)) → U10_AG(y0, membersc20_out_ag([], x0)) we obtained the following new rules [LPAR04]:
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), membersc20_out_ag([], z1))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), membersc20_out_ag([], x2))
(44) Obligation:
Q DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), U17_ag(z1, memberc27_in_ag(z1)))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), U17_ag(x2, memberc27_in_ag(x2)))
U10_AG(z0, membersc20_out_ag([], z2)) → COLOR_MAP1_IN_AG(z0)
U10_AG(.(z0, z1), membersc20_out_ag(x1, x2)) → COLOR_MAP1_IN_AG(.(z0, z1))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), membersc20_out_ag([], z1))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), membersc20_out_ag([], x2))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(45) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule U10_AG(z0, membersc20_out_ag([], z2)) → COLOR_MAP1_IN_AG(z0) we obtained the following new rules [LPAR04]:
U10_AG(.(z0, z1), membersc20_out_ag([], x1)) → COLOR_MAP1_IN_AG(.(z0, z1))
U10_AG(.(z0, z1), membersc20_out_ag([], z1)) → COLOR_MAP1_IN_AG(.(z0, z1))
(46) Obligation:
Q DP problem:The TRS P consists of the following rules:
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), U13_aga(x0, x1, selectc10_in_aga(x1)))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), U17_ag(z1, memberc27_in_ag(z1)))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), U17_ag(x2, memberc27_in_ag(x2)))
U10_AG(.(z0, z1), membersc20_out_ag(x1, x2)) → COLOR_MAP1_IN_AG(.(z0, z1))
U9_AG(.(z0, z1), selectc10_out_aga(z0, .(z0, z1), z1)) → U10_AG(.(z0, z1), membersc20_out_ag([], z1))
U9_AG(.(z0, z1), selectc10_out_aga(x1, .(z0, z1), x2)) → U10_AG(.(z0, z1), membersc20_out_ag([], x2))
U10_AG(.(z0, z1), membersc20_out_ag([], x1)) → COLOR_MAP1_IN_AG(.(z0, z1))
U10_AG(.(z0, z1), membersc20_out_ag([], z1)) → COLOR_MAP1_IN_AG(.(z0, z1))
The TRS R consists of the following rules:
selectc10_in_aga(.(T48, T49)) → selectc10_out_aga(T48, .(T48, T49), T49)
selectc10_in_aga(.(T57, T58)) → U13_aga(T57, T58, selectc10_in_aga(T58))
U13_aga(T57, T58, selectc10_out_aga(T59, T58, X67)) → selectc10_out_aga(T59, .(T57, T58), .(T57, X67))
memberc27_in_ag(.(T105, T106)) → memberc27_out_ag(T105, .(T105, T106))
memberc27_in_ag(.(T114, T115)) → U19_ag(T114, T115, memberc27_in_ag(T115))
U19_ag(T114, T115, memberc27_out_ag(T116, T115)) → memberc27_out_ag(T116, .(T114, T115))
membersc20_in_ag(T85) → U17_ag(T85, memberc27_in_ag(T85))
U17_ag(T85, memberc27_out_ag(T86, T85)) → U18_ag(T86, T85, membersc20_in_ag(T85))
membersc20_in_ag(T126) → membersc20_out_ag([], T126)
U18_ag(T86, T85, membersc20_out_ag(T92, T85)) → membersc20_out_ag(.(T86, T92), T85)
The set Q consists of the following terms:
selectc10_in_aga(x0)
U13_aga(x0, x1, x2)
memberc27_in_ag(x0)
U19_ag(x0, x1, x2)
membersc20_in_ag(x0)
U17_ag(x0, x1)
U18_ag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(47) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.Found a loop by narrowing to the left:
s = U9_AG(.(z0', z1'), selectc10_out_aga(z0', .(z0', z1'), z1')) evaluates to t =U9_AG(.(z0', z1'), selectc10_out_aga(z0', .(z0', z1'), z1'))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequence
U9_AG(.(z0', z1'), selectc10_out_aga(z0', .(z0', z1'), z1')) → U10_AG(.(z0', z1'), membersc20_out_ag([], z1'))
with rule U9_AG(.(z0'', z1''), selectc10_out_aga(z0'', .(z0'', z1''), z1'')) → U10_AG(.(z0'', z1''), membersc20_out_ag([], z1'')) at position [] and matcher [z0'' / z0', z1'' / z1']
U10_AG(.(z0', z1'), membersc20_out_ag([], z1')) → COLOR_MAP1_IN_AG(.(z0', z1'))
with rule U10_AG(.(z0, z1), membersc20_out_ag(x1', x2)) → COLOR_MAP1_IN_AG(.(z0, z1)) at position [] and matcher [z0 / z0', z1 / z1', x1' / [], x2 / z1']
COLOR_MAP1_IN_AG(.(z0', z1')) → U9_AG(.(z0', z1'), selectc10_out_aga(z0', .(z0', z1'), z1'))
with rule COLOR_MAP1_IN_AG(.(x0, x1)) → U9_AG(.(x0, x1), selectc10_out_aga(x0, .(x0, x1), x1))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.