(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_AAREVERSE10_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_AA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE10_IN_AA to REVERSE10_IN_AA.



(27) NO