Left Termination of the query pattern in_order_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

in_order(void, []).
in_order(tree(X, Left, Right), Xs) :- ','(in_order(Left, Ls), ','(in_order(Right, Rs), app(Ls, .(X, Rs), Xs))).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

in_order(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)


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

IN_ORDER_IN(tree(X, Left, Right), Xs) → U11(X, Left, Right, Xs, in_order_in(Left, Ls))
IN_ORDER_IN(tree(X, Left, Right), Xs) → IN_ORDER_IN(Left, Ls)
U11(X, Left, Right, Xs, in_order_out(Left, Ls)) → U21(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U11(X, Left, Right, Xs, in_order_out(Left, Ls)) → IN_ORDER_IN(Right, Rs)
U21(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U31(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
U21(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → APP_IN(Ls, .(X, Rs), Xs)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U41(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)
U41(x1, x2, x3, x4, x5)  =  U41(x1, x5)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x5, x6)
IN_ORDER_IN(x1, x2)  =  IN_ORDER_IN(x1)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x3, x5)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

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

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

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

IN_ORDER_IN(tree(X, Left, Right), Xs) → U11(X, Left, Right, Xs, in_order_in(Left, Ls))
IN_ORDER_IN(tree(X, Left, Right), Xs) → IN_ORDER_IN(Left, Ls)
U11(X, Left, Right, Xs, in_order_out(Left, Ls)) → U21(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U11(X, Left, Right, Xs, in_order_out(Left, Ls)) → IN_ORDER_IN(Right, Rs)
U21(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U31(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
U21(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → APP_IN(Ls, .(X, Rs), Xs)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U41(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)
U41(x1, x2, x3, x4, x5)  =  U41(x1, x5)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x5, x6)
IN_ORDER_IN(x1, x2)  =  IN_ORDER_IN(x1)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x3, x5)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

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

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

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

The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

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

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

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

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

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

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

APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)

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
                ↳ PiDPToQDPProof

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

IN_ORDER_IN(tree(X, Left, Right), Xs) → IN_ORDER_IN(Left, Ls)
U11(X, Left, Right, Xs, in_order_out(Left, Ls)) → IN_ORDER_IN(Right, Rs)
IN_ORDER_IN(tree(X, Left, Right), Xs) → U11(X, Left, Right, Xs, in_order_in(Left, Ls))

The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right), Xs) → U1(X, Left, Right, Xs, in_order_in(Left, Ls))
in_order_in(void, []) → in_order_out(void, [])
U1(X, Left, Right, Xs, in_order_out(Left, Ls)) → U2(X, Left, Right, Xs, Ls, in_order_in(Right, Rs))
U2(X, Left, Right, Xs, Ls, in_order_out(Right, Rs)) → U3(X, Left, Right, Xs, app_in(Ls, .(X, Rs), Xs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U4(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U4(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Left, Right, Xs, app_out(Ls, .(X, Rs), Xs)) → in_order_out(tree(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_in(x1, x2)  =  in_order_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x3, x5)
void  =  void
[]  =  []
in_order_out(x1, x2)  =  in_order_out(x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x5, x6)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
app_in(x1, x2, x3)  =  app_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
app_out(x1, x2, x3)  =  app_out(x3)
IN_ORDER_IN(x1, x2)  =  IN_ORDER_IN(x1)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x3, x5)

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
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ QDPSizeChangeProof

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

IN_ORDER_IN(tree(X, Left, Right)) → U11(X, Right, in_order_in(Left))
U11(X, Right, in_order_out(Ls)) → IN_ORDER_IN(Right)
IN_ORDER_IN(tree(X, Left, Right)) → IN_ORDER_IN(Left)

The TRS R consists of the following rules:

in_order_in(tree(X, Left, Right)) → U1(X, Right, in_order_in(Left))
in_order_in(void) → in_order_out([])
U1(X, Right, in_order_out(Ls)) → U2(X, Ls, in_order_in(Right))
U2(X, Ls, in_order_out(Rs)) → U3(app_in(Ls, .(X, Rs)))
app_in(.(X, Xs), Ys) → U4(X, app_in(Xs, Ys))
app_in([], X) → app_out(X)
U4(X, app_out(Zs)) → app_out(.(X, Zs))
U3(app_out(Xs)) → in_order_out(Xs)

The set Q consists of the following terms:

in_order_in(x0)
U1(x0, x1, x2)
U2(x0, x1, x2)
app_in(x0, x1)
U4(x0, x1)
U3(x0)

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: