Left Termination of the query pattern reverse_in_2(a, g) w.r.t. the given Prolog program could not be shown:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔)
        ↳20 TRUE
      ↳21 PiDP
        ↳22 UsableRulesProof (⇔)
        ↳23 PiDP
        ↳24 PiDPToQDPProof (⇐)
        ↳25 QDP
        ↳26 NonTerminationProof (⇔)
        ↳27 FALSE
  ↳28 PrologToPiTRSProof (⇐)
    ↳29 PiTRS
    ↳30 DependencyPairsProof (⇔)
    ↳31 PiDP
    ↳32 DependencyGraphProof (⇔)
    ↳33 AND
      ↳34 PiDP
        ↳35 UsableRulesProof (⇔)
        ↳36 PiDP
        ↳37 PiDPToQDPProof (⇐)
        ↳38 QDP
        ↳39 QDPSizeChangeProof (⇔)
        ↳40 TRUE
      ↳41 PiDP
        ↳42 UsableRulesProof (⇔)
        ↳43 PiDP
        ↳44 PiDPToQDPProof (⇐)
        ↳45 QDP
        ↳46 QDPSizeChangeProof (⇔)
        ↳47 TRUE
      ↳48 PiDP
        ↳49 UsableRulesProof (⇔)
        ↳50 PiDP
        ↳51 PiDPToQDPProof (⇐)
        ↳52 QDP
        ↳53 NonTerminationProof (⇔)
        ↳54 FALSE

(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) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (f,b) (f,f)
app_in: (b,b,f) (b,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)

(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:

REVERSE_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AG(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
REVERSE_IN_AA(.(X, Xs), Ys) → U2_AA(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AG(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGG(Zs, .(X, []), Ys)
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → U1_GGG(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x2, x4)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x5)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x2, x4)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2, x3, x4, x5)  =  U1_GGG(x5)

We have to consider all (P,R,Pi)-chains

(4) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AG(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
REVERSE_IN_AA(.(X, Xs), Ys) → U2_AA(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AG(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGG(Zs, .(X, []), Ys)
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → U1_GGG(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x2, x4)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x5)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x2, x4)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2, x3, x4, x5)  =  U1_GGG(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 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, 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:

APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, 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:

APP_IN_GGG(.(Xs), Ys, .(Zs)) → APP_IN_GGG(Xs, Ys, Zs)

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:

  • APP_IN_GGG(.(Xs), Ys, .(Zs)) → APP_IN_GGG(Xs, Ys, Zs)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 > 3

(13) TRUE

(14) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

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:

APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

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:

APP_IN_GGA(.(Xs), Ys) → APP_IN_GGA(Xs, Ys)

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:

  • APP_IN_GGA(.(Xs), Ys) → APP_IN_GGA(Xs, Ys)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AA(x1, x2)  =  REVERSE_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:

REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
REVERSE_IN_AA(x1, x2)  =  REVERSE_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:

REVERSE_IN_AA → REVERSE_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 = REVERSE_IN_AA evaluates to t =REVERSE_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 REVERSE_IN_AA to REVERSE_IN_AA.



(27) FALSE

(28) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (f,b) (f,f)
app_in: (b,b,f) (b,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(29) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)

(30) 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:

REVERSE_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AG(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
REVERSE_IN_AA(.(X, Xs), Ys) → U2_AA(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AG(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGG(Zs, .(X, []), Ys)
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → U1_GGG(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x2, x4)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x2, x3, x5)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x2, x3, x4)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2, x3, x4, x5)  =  U1_GGG(x2, x3, x4, x5)

We have to consider all (P,R,Pi)-chains

(31) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AG(.(X, Xs), Ys) → U2_AG(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AG(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
REVERSE_IN_AA(.(X, Xs), Ys) → U2_AA(X, Xs, Ys, reverse_in_aa(Xs, Zs))
REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_AA(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_AG(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
U2_AG(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → APP_IN_GGG(Zs, .(X, []), Ys)
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → U1_GGG(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x2, x4)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x2, x3, x5)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x2, x3, x4)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)
U1_GGG(x1, x2, x3, x4, x5)  =  U1_GGG(x2, x3, x4, x5)

We have to consider all (P,R,Pi)-chains

(32) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 9 less nodes.

(33) Complex Obligation (AND)

(34) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)

We have to consider all (P,R,Pi)-chains

(35) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(36) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGG(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP_IN_GGG(x1, x2, x3)  =  APP_IN_GGG(x1, x2, x3)

We have to consider all (P,R,Pi)-chains

(37) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP_IN_GGG(.(Xs), Ys, .(Zs)) → APP_IN_GGG(Xs, Ys, Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(39) 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:

  • APP_IN_GGG(.(Xs), Ys, .(Zs)) → APP_IN_GGG(Xs, Ys, Zs)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 > 3

(40) TRUE

(41) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains

(42) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(43) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains

(44) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(45) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP_IN_GGA(.(Xs), Ys) → APP_IN_GGA(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(46) 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:

  • APP_IN_GGA(.(Xs), Ys) → APP_IN_GGA(Xs, Ys)
    The graph contains the following edges 1 > 1, 2 >= 2

(47) TRUE

(48) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)

The TRS R consists of the following rules:

reverse_in_ag(.(X, Xs), Ys) → U2_ag(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa(.(X, Xs), Ys) → U2_aa(X, Xs, Ys, reverse_in_aa(Xs, Zs))
reverse_in_aa([], []) → reverse_out_aa([], [])
U2_aa(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_aa(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_aa(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_aa(.(X, Xs), Ys)
U2_ag(X, Xs, Ys, reverse_out_aa(Xs, Zs)) → U3_ag(X, Xs, Ys, app_in_ggg(Zs, .(X, []), Ys))
app_in_ggg(.(X, Xs), Ys, .(X, Zs)) → U1_ggg(X, Xs, Ys, Zs, app_in_ggg(Xs, Ys, Zs))
app_in_ggg([], Ys, Ys) → app_out_ggg([], Ys, Ys)
U1_ggg(X, Xs, Ys, Zs, app_out_ggg(Xs, Ys, Zs)) → app_out_ggg(.(X, Xs), Ys, .(X, Zs))
U3_ag(X, Xs, Ys, app_out_ggg(Zs, .(X, []), Ys)) → reverse_out_ag(.(X, Xs), Ys)
reverse_in_ag([], []) → reverse_out_ag([], [])

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
reverse_in_aa(x1, x2)  =  reverse_in_aa
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
reverse_out_aa(x1, x2)  =  reverse_out_aa(x1, x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x2, x4)
app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
.(x1, x2)  =  .(x2)
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x2, x3, x5)
[]  =  []
app_out_gga(x1, x2, x3)  =  app_out_gga(x1, x2, x3)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x2, x3, x4)
app_in_ggg(x1, x2, x3)  =  app_in_ggg(x1, x2, x3)
U1_ggg(x1, x2, x3, x4, x5)  =  U1_ggg(x2, x3, x4, x5)
app_out_ggg(x1, x2, x3)  =  app_out_ggg(x1, x2, x3)
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA

We have to consider all (P,R,Pi)-chains

(49) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(50) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AA(.(X, Xs), Ys) → REVERSE_IN_AA(Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
REVERSE_IN_AA(x1, x2)  =  REVERSE_IN_AA

We have to consider all (P,R,Pi)-chains

(51) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(52) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REVERSE_IN_AA → REVERSE_IN_AA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(53) 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 = REVERSE_IN_AA evaluates to t =REVERSE_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 REVERSE_IN_AA to REVERSE_IN_AA.



(54) FALSE