(0) Obligation:
Clauses:
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
app([], Ys, Ys).
reverse(.(X, Xs), Ys) :- ','(reverse(Xs, Zs), app(Zs, .(X, []), Ys)).
reverse([], []).
Queries:
reverse(a,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:
reverse10(.(T42, T41), X69) :- reverse10(T41, X68).
reverse10(.(T46, T41), X69) :- ','(reversec10(T41, T45), app18(T45, T46, X69)).
app18(.(T64, T67), T68, .(T64, X112)) :- app18(T67, T68, X112).
app31(.(T103, T107), T108, .(T103, T106)) :- app31(T107, T108, T106).
reverse1(.(T23, .(T22, T21)), T8) :- reverse10(T21, X32).
reverse1(.(T28, .(T27, T21)), T8) :- ','(reversec10(T21, T26), app18(T26, T27, X33)).
reverse1(.(T79, .(T27, T21)), T8) :- ','(reversec10(T21, T26), ','(appc18(T26, T27, T78), app31(T78, T79, T8))).
Clauses:
reversec10(.(T46, T41), X69) :- ','(reversec10(T41, T45), appc18(T45, T46, X69)).
reversec10([], []).
appc18(.(T64, T67), T68, .(T64, X112)) :- appc18(T67, T68, X112).
appc18([], T74, .(T74, [])).
appc31(.(T103, T107), T108, .(T103, T106)) :- appc31(T107, T108, T106).
appc31([], T115, .(T115, [])).
Afs:
reverse1(x1, x2) = reverse1(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:
reverse1_in: (f,b)
reverse10_in: (f,f)
reversec10_in: (f,f)
appc18_in: (b,f,f)
app18_in: (b,f,f)
app31_in: (b,f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE1_IN_AG(.(T23, .(T22, T21)), T8) → U6_AG(T23, T22, T21, T8, reverse10_in_aa(T21, X32))
REVERSE1_IN_AG(.(T23, .(T22, T21)), T8) → REVERSE10_IN_AA(T21, X32)
REVERSE10_IN_AA(.(T42, T41), X69) → U1_AA(T42, T41, X69, reverse10_in_aa(T41, X68))
REVERSE10_IN_AA(.(T42, T41), X69) → REVERSE10_IN_AA(T41, X68)
REVERSE10_IN_AA(.(T46, T41), X69) → U2_AA(T46, T41, X69, reversec10_in_aa(T41, T45))
U2_AA(T46, T41, X69, reversec10_out_aa(T41, T45)) → U3_AA(T46, T41, X69, app18_in_gaa(T45, T46, X69))
U2_AA(T46, T41, X69, reversec10_out_aa(T41, T45)) → APP18_IN_GAA(T45, T46, X69)
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → U4_GAA(T64, T67, T68, X112, app18_in_gaa(T67, T68, X112))
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → APP18_IN_GAA(T67, T68, X112)
REVERSE1_IN_AG(.(T28, .(T27, T21)), T8) → U7_AG(T28, T27, T21, T8, reversec10_in_aa(T21, T26))
U7_AG(T28, T27, T21, T8, reversec10_out_aa(T21, T26)) → U8_AG(T28, T27, T21, T8, app18_in_gaa(T26, T27, X33))
U7_AG(T28, T27, T21, T8, reversec10_out_aa(T21, T26)) → APP18_IN_GAA(T26, T27, X33)
REVERSE1_IN_AG(.(T79, .(T27, T21)), T8) → U9_AG(T79, T27, T21, T8, reversec10_in_aa(T21, T26))
U9_AG(T79, T27, T21, T8, reversec10_out_aa(T21, T26)) → U10_AG(T79, T27, T21, T8, appc18_in_gaa(T26, T27, T78))
U10_AG(T79, T27, T21, T8, appc18_out_gaa(T26, T27, T78)) → U11_AG(T79, T27, T21, T8, app31_in_gag(T78, T79, T8))
U10_AG(T79, T27, T21, T8, appc18_out_gaa(T26, T27, T78)) → APP31_IN_GAG(T78, T79, T8)
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → U5_GAG(T103, T107, T108, T106, app31_in_gag(T107, T108, T106))
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → APP31_IN_GAG(T107, T108, T106)
The TRS R consists of the following rules:
reversec10_in_aa(.(T46, T41), X69) → U13_aa(T46, T41, X69, reversec10_in_aa(T41, T45))
reversec10_in_aa([], []) → reversec10_out_aa([], [])
U13_aa(T46, T41, X69, reversec10_out_aa(T41, T45)) → U14_aa(T46, T41, X69, appc18_in_gaa(T45, T46, X69))
appc18_in_gaa(.(T64, T67), T68, .(T64, X112)) → U15_gaa(T64, T67, T68, X112, appc18_in_gaa(T67, T68, X112))
appc18_in_gaa([], T74, .(T74, [])) → appc18_out_gaa([], T74, .(T74, []))
U15_gaa(T64, T67, T68, X112, appc18_out_gaa(T67, T68, X112)) → appc18_out_gaa(.(T64, T67), T68, .(T64, X112))
U14_aa(T46, T41, X69, appc18_out_gaa(T45, T46, X69)) → reversec10_out_aa(.(T46, T41), X69)
The argument filtering Pi contains the following mapping:
reverse10_in_aa(
x1,
x2) =
reverse10_in_aa
reversec10_in_aa(
x1,
x2) =
reversec10_in_aa
U13_aa(
x1,
x2,
x3,
x4) =
U13_aa(
x4)
reversec10_out_aa(
x1,
x2) =
reversec10_out_aa(
x1,
x2)
U14_aa(
x1,
x2,
x3,
x4) =
U14_aa(
x2,
x4)
appc18_in_gaa(
x1,
x2,
x3) =
appc18_in_gaa(
x1)
.(
x1,
x2) =
.(
x2)
U15_gaa(
x1,
x2,
x3,
x4,
x5) =
U15_gaa(
x2,
x5)
[] =
[]
appc18_out_gaa(
x1,
x2,
x3) =
appc18_out_gaa(
x1,
x3)
app18_in_gaa(
x1,
x2,
x3) =
app18_in_gaa(
x1)
app31_in_gag(
x1,
x2,
x3) =
app31_in_gag(
x1,
x3)
REVERSE1_IN_AG(
x1,
x2) =
REVERSE1_IN_AG(
x2)
U6_AG(
x1,
x2,
x3,
x4,
x5) =
U6_AG(
x4,
x5)
REVERSE10_IN_AA(
x1,
x2) =
REVERSE10_IN_AA
U1_AA(
x1,
x2,
x3,
x4) =
U1_AA(
x4)
U2_AA(
x1,
x2,
x3,
x4) =
U2_AA(
x4)
U3_AA(
x1,
x2,
x3,
x4) =
U3_AA(
x2,
x4)
APP18_IN_GAA(
x1,
x2,
x3) =
APP18_IN_GAA(
x1)
U4_GAA(
x1,
x2,
x3,
x4,
x5) =
U4_GAA(
x2,
x5)
U7_AG(
x1,
x2,
x3,
x4,
x5) =
U7_AG(
x4,
x5)
U8_AG(
x1,
x2,
x3,
x4,
x5) =
U8_AG(
x3,
x4,
x5)
U9_AG(
x1,
x2,
x3,
x4,
x5) =
U9_AG(
x4,
x5)
U10_AG(
x1,
x2,
x3,
x4,
x5) =
U10_AG(
x3,
x4,
x5)
U11_AG(
x1,
x2,
x3,
x4,
x5) =
U11_AG(
x3,
x4,
x5)
APP31_IN_GAG(
x1,
x2,
x3) =
APP31_IN_GAG(
x1,
x3)
U5_GAG(
x1,
x2,
x3,
x4,
x5) =
U5_GAG(
x2,
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:
REVERSE1_IN_AG(.(T23, .(T22, T21)), T8) → U6_AG(T23, T22, T21, T8, reverse10_in_aa(T21, X32))
REVERSE1_IN_AG(.(T23, .(T22, T21)), T8) → REVERSE10_IN_AA(T21, X32)
REVERSE10_IN_AA(.(T42, T41), X69) → U1_AA(T42, T41, X69, reverse10_in_aa(T41, X68))
REVERSE10_IN_AA(.(T42, T41), X69) → REVERSE10_IN_AA(T41, X68)
REVERSE10_IN_AA(.(T46, T41), X69) → U2_AA(T46, T41, X69, reversec10_in_aa(T41, T45))
U2_AA(T46, T41, X69, reversec10_out_aa(T41, T45)) → U3_AA(T46, T41, X69, app18_in_gaa(T45, T46, X69))
U2_AA(T46, T41, X69, reversec10_out_aa(T41, T45)) → APP18_IN_GAA(T45, T46, X69)
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → U4_GAA(T64, T67, T68, X112, app18_in_gaa(T67, T68, X112))
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → APP18_IN_GAA(T67, T68, X112)
REVERSE1_IN_AG(.(T28, .(T27, T21)), T8) → U7_AG(T28, T27, T21, T8, reversec10_in_aa(T21, T26))
U7_AG(T28, T27, T21, T8, reversec10_out_aa(T21, T26)) → U8_AG(T28, T27, T21, T8, app18_in_gaa(T26, T27, X33))
U7_AG(T28, T27, T21, T8, reversec10_out_aa(T21, T26)) → APP18_IN_GAA(T26, T27, X33)
REVERSE1_IN_AG(.(T79, .(T27, T21)), T8) → U9_AG(T79, T27, T21, T8, reversec10_in_aa(T21, T26))
U9_AG(T79, T27, T21, T8, reversec10_out_aa(T21, T26)) → U10_AG(T79, T27, T21, T8, appc18_in_gaa(T26, T27, T78))
U10_AG(T79, T27, T21, T8, appc18_out_gaa(T26, T27, T78)) → U11_AG(T79, T27, T21, T8, app31_in_gag(T78, T79, T8))
U10_AG(T79, T27, T21, T8, appc18_out_gaa(T26, T27, T78)) → APP31_IN_GAG(T78, T79, T8)
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → U5_GAG(T103, T107, T108, T106, app31_in_gag(T107, T108, T106))
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → APP31_IN_GAG(T107, T108, T106)
The TRS R consists of the following rules:
reversec10_in_aa(.(T46, T41), X69) → U13_aa(T46, T41, X69, reversec10_in_aa(T41, T45))
reversec10_in_aa([], []) → reversec10_out_aa([], [])
U13_aa(T46, T41, X69, reversec10_out_aa(T41, T45)) → U14_aa(T46, T41, X69, appc18_in_gaa(T45, T46, X69))
appc18_in_gaa(.(T64, T67), T68, .(T64, X112)) → U15_gaa(T64, T67, T68, X112, appc18_in_gaa(T67, T68, X112))
appc18_in_gaa([], T74, .(T74, [])) → appc18_out_gaa([], T74, .(T74, []))
U15_gaa(T64, T67, T68, X112, appc18_out_gaa(T67, T68, X112)) → appc18_out_gaa(.(T64, T67), T68, .(T64, X112))
U14_aa(T46, T41, X69, appc18_out_gaa(T45, T46, X69)) → reversec10_out_aa(.(T46, T41), X69)
The argument filtering Pi contains the following mapping:
reverse10_in_aa(
x1,
x2) =
reverse10_in_aa
reversec10_in_aa(
x1,
x2) =
reversec10_in_aa
U13_aa(
x1,
x2,
x3,
x4) =
U13_aa(
x4)
reversec10_out_aa(
x1,
x2) =
reversec10_out_aa(
x1,
x2)
U14_aa(
x1,
x2,
x3,
x4) =
U14_aa(
x2,
x4)
appc18_in_gaa(
x1,
x2,
x3) =
appc18_in_gaa(
x1)
.(
x1,
x2) =
.(
x2)
U15_gaa(
x1,
x2,
x3,
x4,
x5) =
U15_gaa(
x2,
x5)
[] =
[]
appc18_out_gaa(
x1,
x2,
x3) =
appc18_out_gaa(
x1,
x3)
app18_in_gaa(
x1,
x2,
x3) =
app18_in_gaa(
x1)
app31_in_gag(
x1,
x2,
x3) =
app31_in_gag(
x1,
x3)
REVERSE1_IN_AG(
x1,
x2) =
REVERSE1_IN_AG(
x2)
U6_AG(
x1,
x2,
x3,
x4,
x5) =
U6_AG(
x4,
x5)
REVERSE10_IN_AA(
x1,
x2) =
REVERSE10_IN_AA
U1_AA(
x1,
x2,
x3,
x4) =
U1_AA(
x4)
U2_AA(
x1,
x2,
x3,
x4) =
U2_AA(
x4)
U3_AA(
x1,
x2,
x3,
x4) =
U3_AA(
x2,
x4)
APP18_IN_GAA(
x1,
x2,
x3) =
APP18_IN_GAA(
x1)
U4_GAA(
x1,
x2,
x3,
x4,
x5) =
U4_GAA(
x2,
x5)
U7_AG(
x1,
x2,
x3,
x4,
x5) =
U7_AG(
x4,
x5)
U8_AG(
x1,
x2,
x3,
x4,
x5) =
U8_AG(
x3,
x4,
x5)
U9_AG(
x1,
x2,
x3,
x4,
x5) =
U9_AG(
x4,
x5)
U10_AG(
x1,
x2,
x3,
x4,
x5) =
U10_AG(
x3,
x4,
x5)
U11_AG(
x1,
x2,
x3,
x4,
x5) =
U11_AG(
x3,
x4,
x5)
APP31_IN_GAG(
x1,
x2,
x3) =
APP31_IN_GAG(
x1,
x3)
U5_GAG(
x1,
x2,
x3,
x4,
x5) =
U5_GAG(
x2,
x4,
x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 15 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → APP31_IN_GAG(T107, T108, T106)
The TRS R consists of the following rules:
reversec10_in_aa(.(T46, T41), X69) → U13_aa(T46, T41, X69, reversec10_in_aa(T41, T45))
reversec10_in_aa([], []) → reversec10_out_aa([], [])
U13_aa(T46, T41, X69, reversec10_out_aa(T41, T45)) → U14_aa(T46, T41, X69, appc18_in_gaa(T45, T46, X69))
appc18_in_gaa(.(T64, T67), T68, .(T64, X112)) → U15_gaa(T64, T67, T68, X112, appc18_in_gaa(T67, T68, X112))
appc18_in_gaa([], T74, .(T74, [])) → appc18_out_gaa([], T74, .(T74, []))
U15_gaa(T64, T67, T68, X112, appc18_out_gaa(T67, T68, X112)) → appc18_out_gaa(.(T64, T67), T68, .(T64, X112))
U14_aa(T46, T41, X69, appc18_out_gaa(T45, T46, X69)) → reversec10_out_aa(.(T46, T41), X69)
The argument filtering Pi contains the following mapping:
reversec10_in_aa(
x1,
x2) =
reversec10_in_aa
U13_aa(
x1,
x2,
x3,
x4) =
U13_aa(
x4)
reversec10_out_aa(
x1,
x2) =
reversec10_out_aa(
x1,
x2)
U14_aa(
x1,
x2,
x3,
x4) =
U14_aa(
x2,
x4)
appc18_in_gaa(
x1,
x2,
x3) =
appc18_in_gaa(
x1)
.(
x1,
x2) =
.(
x2)
U15_gaa(
x1,
x2,
x3,
x4,
x5) =
U15_gaa(
x2,
x5)
[] =
[]
appc18_out_gaa(
x1,
x2,
x3) =
appc18_out_gaa(
x1,
x3)
APP31_IN_GAG(
x1,
x2,
x3) =
APP31_IN_GAG(
x1,
x3)
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:
APP31_IN_GAG(.(T103, T107), T108, .(T103, T106)) → APP31_IN_GAG(T107, T108, T106)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x2)
APP31_IN_GAG(
x1,
x2,
x3) =
APP31_IN_GAG(
x1,
x3)
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:
APP31_IN_GAG(.(T107), .(T106)) → APP31_IN_GAG(T107, T106)
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:
- APP31_IN_GAG(.(T107), .(T106)) → APP31_IN_GAG(T107, T106)
The graph contains the following edges 1 > 1, 2 > 2
(13) YES
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → APP18_IN_GAA(T67, T68, X112)
The TRS R consists of the following rules:
reversec10_in_aa(.(T46, T41), X69) → U13_aa(T46, T41, X69, reversec10_in_aa(T41, T45))
reversec10_in_aa([], []) → reversec10_out_aa([], [])
U13_aa(T46, T41, X69, reversec10_out_aa(T41, T45)) → U14_aa(T46, T41, X69, appc18_in_gaa(T45, T46, X69))
appc18_in_gaa(.(T64, T67), T68, .(T64, X112)) → U15_gaa(T64, T67, T68, X112, appc18_in_gaa(T67, T68, X112))
appc18_in_gaa([], T74, .(T74, [])) → appc18_out_gaa([], T74, .(T74, []))
U15_gaa(T64, T67, T68, X112, appc18_out_gaa(T67, T68, X112)) → appc18_out_gaa(.(T64, T67), T68, .(T64, X112))
U14_aa(T46, T41, X69, appc18_out_gaa(T45, T46, X69)) → reversec10_out_aa(.(T46, T41), X69)
The argument filtering Pi contains the following mapping:
reversec10_in_aa(
x1,
x2) =
reversec10_in_aa
U13_aa(
x1,
x2,
x3,
x4) =
U13_aa(
x4)
reversec10_out_aa(
x1,
x2) =
reversec10_out_aa(
x1,
x2)
U14_aa(
x1,
x2,
x3,
x4) =
U14_aa(
x2,
x4)
appc18_in_gaa(
x1,
x2,
x3) =
appc18_in_gaa(
x1)
.(
x1,
x2) =
.(
x2)
U15_gaa(
x1,
x2,
x3,
x4,
x5) =
U15_gaa(
x2,
x5)
[] =
[]
appc18_out_gaa(
x1,
x2,
x3) =
appc18_out_gaa(
x1,
x3)
APP18_IN_GAA(
x1,
x2,
x3) =
APP18_IN_GAA(
x1)
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:
APP18_IN_GAA(.(T64, T67), T68, .(T64, X112)) → APP18_IN_GAA(T67, T68, X112)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x2)
APP18_IN_GAA(
x1,
x2,
x3) =
APP18_IN_GAA(
x1)
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:
APP18_IN_GAA(.(T67)) → APP18_IN_GAA(T67)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(19) 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:
- APP18_IN_GAA(.(T67)) → APP18_IN_GAA(T67)
The graph contains the following edges 1 > 1
(20) YES
(21) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE10_IN_AA(.(T42, T41), X69) → REVERSE10_IN_AA(T41, X68)
The TRS R consists of the following rules:
reversec10_in_aa(.(T46, T41), X69) → U13_aa(T46, T41, X69, reversec10_in_aa(T41, T45))
reversec10_in_aa([], []) → reversec10_out_aa([], [])
U13_aa(T46, T41, X69, reversec10_out_aa(T41, T45)) → U14_aa(T46, T41, X69, appc18_in_gaa(T45, T46, X69))
appc18_in_gaa(.(T64, T67), T68, .(T64, X112)) → U15_gaa(T64, T67, T68, X112, appc18_in_gaa(T67, T68, X112))
appc18_in_gaa([], T74, .(T74, [])) → appc18_out_gaa([], T74, .(T74, []))
U15_gaa(T64, T67, T68, X112, appc18_out_gaa(T67, T68, X112)) → appc18_out_gaa(.(T64, T67), T68, .(T64, X112))
U14_aa(T46, T41, X69, appc18_out_gaa(T45, T46, X69)) → reversec10_out_aa(.(T46, T41), X69)
The argument filtering Pi contains the following mapping:
reversec10_in_aa(
x1,
x2) =
reversec10_in_aa
U13_aa(
x1,
x2,
x3,
x4) =
U13_aa(
x4)
reversec10_out_aa(
x1,
x2) =
reversec10_out_aa(
x1,
x2)
U14_aa(
x1,
x2,
x3,
x4) =
U14_aa(
x2,
x4)
appc18_in_gaa(
x1,
x2,
x3) =
appc18_in_gaa(
x1)
.(
x1,
x2) =
.(
x2)
U15_gaa(
x1,
x2,
x3,
x4,
x5) =
U15_gaa(
x2,
x5)
[] =
[]
appc18_out_gaa(
x1,
x2,
x3) =
appc18_out_gaa(
x1,
x3)
REVERSE10_IN_AA(
x1,
x2) =
REVERSE10_IN_AA
We have to consider all (P,R,Pi)-chains
(22) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(23) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE10_IN_AA(.(T42, T41), X69) → REVERSE10_IN_AA(T41, X68)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x2)
REVERSE10_IN_AA(
x1,
x2) =
REVERSE10_IN_AA
We have to consider all (P,R,Pi)-chains
(24) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REVERSE10_IN_AA → REVERSE10_IN_AA
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(26) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
REVERSE10_IN_AA evaluates to t =
REVERSE10_IN_AAThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from REVERSE10_IN_AA to REVERSE10_IN_AA.
(27) NO