Left Termination of the query pattern rev_in_2(a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

rev([], []).
rev(.(X, Xs), Ys) :- ','(rev(Xs, Zs), app(Zs, .(X, []), Ys)).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

rev(a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
rev_in: (f,b) (f,f)
app_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REV_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AG(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
REV_IN_AA(.(X, Xs), Ys) → U1_AA(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AA(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AA(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAA(Zs, .(X, []), Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AG(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAG(Zs, .(X, []), Ys)
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U3_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
REV_IN_AA(x1, x2)  =  REV_IN_AA
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
REV_IN_AG(x1, x2)  =  REV_IN_AG(x2)
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x5)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U1_AA(x1, x2, x3, x4)  =  U1_AA(x4)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

REV_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AG(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
REV_IN_AA(.(X, Xs), Ys) → U1_AA(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AA(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AA(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAA(Zs, .(X, []), Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AG(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAG(Zs, .(X, []), Ys)
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U3_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
REV_IN_AA(x1, x2)  =  REV_IN_AA
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
REV_IN_AG(x1, x2)  =  REV_IN_AG(x2)
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x5)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U1_AA(x1, x2, x3, x4)  =  U1_AA(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 9 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APP_IN_AAG(.(X, Zs)) → APP_IN_AAG(Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APP_IN_AAA → APP_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN_AAA → APP_IN_AAA

The TRS R consists of the following rules:none


s = APP_IN_AAA evaluates to t =APP_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

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





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
REV_IN_AA(x1, x2)  =  REV_IN_AA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

REV_IN_AA → REV_IN_AA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

REV_IN_AA → REV_IN_AA

The TRS R consists of the following rules:none


s = REV_IN_AA evaluates to t =REV_IN_AA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

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




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
rev_in: (f,b) (f,f)
app_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REV_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AG(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
REV_IN_AA(.(X, Xs), Ys) → U1_AA(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AA(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AA(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAA(Zs, .(X, []), Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AG(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAG(Zs, .(X, []), Ys)
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U3_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
REV_IN_AA(x1, x2)  =  REV_IN_AA
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
REV_IN_AG(x1, x2)  =  REV_IN_AG(x2)
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x4, x5)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U1_AA(x1, x2, x3, x4)  =  U1_AA(x4)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

REV_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AG(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
REV_IN_AA(.(X, Xs), Ys) → U1_AA(X, Xs, Ys, rev_in_aa(Xs, Zs))
REV_IN_AA(.(X, Xs), Ys) → REV_IN_AA(Xs, Zs)
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AA(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
U1_AA(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAA(Zs, .(X, []), Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_AG(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
U1_AG(X, Xs, Ys, rev_out_aa(Xs, Zs)) → APP_IN_AAG(Zs, .(X, []), Ys)
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U3_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
REV_IN_AA(x1, x2)  =  REV_IN_AA
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
REV_IN_AG(x1, x2)  =  REV_IN_AG(x2)
U2_AA(x1, x2, x3, x4)  =  U2_AA(x4)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x4, x5)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U1_AA(x1, x2, x3, x4)  =  U1_AA(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 9 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)
APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

APP_IN_AAG(.(X, Zs)) → APP_IN_AAG(Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

APP_IN_AAA → APP_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN_AAA → APP_IN_AAA

The TRS R consists of the following rules:none


s = APP_IN_AAA evaluates to t =APP_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

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





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

rev_in_ag([], []) → rev_out_ag([], [])
rev_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, rev_in_aa(Xs, Zs))
rev_in_aa([], []) → rev_out_aa([], [])
rev_in_aa(.(X, Xs), Ys) → U1_aa(X, Xs, Ys, rev_in_aa(Xs, Zs))
U1_aa(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_aa(X, Xs, Ys, app_in_aaa(Zs, .(X, []), Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_aa(X, Xs, Ys, app_out_aaa(Zs, .(X, []), Ys)) → rev_out_aa(.(X, Xs), Ys)
U1_ag(X, Xs, Ys, rev_out_aa(Xs, Zs)) → U2_ag(X, Xs, Ys, app_in_aag(Zs, .(X, []), Ys))
app_in_aag([], X, X) → app_out_aag([], X, X)
app_in_aag(.(X, Xs), Ys, .(X, Zs)) → U3_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
U3_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) → app_out_aag(.(X, Xs), Ys, .(X, Zs))
U2_ag(X, Xs, Ys, app_out_aag(Zs, .(X, []), Ys)) → rev_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
rev_in_ag(x1, x2)  =  rev_in_ag(x2)
[]  =  []
rev_out_ag(x1, x2)  =  rev_out_ag(x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
rev_in_aa(x1, x2)  =  rev_in_aa
rev_out_aa(x1, x2)  =  rev_out_aa
U1_aa(x1, x2, x3, x4)  =  U1_aa(x4)
U2_aa(x1, x2, x3, x4)  =  U2_aa(x4)
.(x1, x2)  =  .(x1, x2)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x4, x5)
REV_IN_AA(x1, x2)  =  REV_IN_AA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof

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

REV_IN_AA → REV_IN_AA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

REV_IN_AA → REV_IN_AA

The TRS R consists of the following rules:none


s = REV_IN_AA evaluates to t =REV_IN_AA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

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