Left Termination of the query pattern map_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(val_i, val_j).
map(.(X, Xs), .(Y, Ys)) :- ','(p(X, Y), map(Xs, Ys)).
map([], []).

Queries:

map(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:

map_in([], []) → map_out([], [])
map_in(.(X, Xs), .(Y, Ys)) → U1(X, Xs, Y, Ys, p_in(X, Y))
p_in(val_i, val_j) → p_out(val_i, val_j)
U1(X, Xs, Y, Ys, p_out(X, Y)) → U2(X, Xs, Y, Ys, map_in(Xs, Ys))
U2(X, Xs, Y, Ys, map_out(Xs, Ys)) → map_out(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in(x1, x2)  =  map_in(x1)
[]  =  []
map_out(x1, x2)  =  map_out(x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)

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:

map_in([], []) → map_out([], [])
map_in(.(X, Xs), .(Y, Ys)) → U1(X, Xs, Y, Ys, p_in(X, Y))
p_in(val_i, val_j) → p_out(val_i, val_j)
U1(X, Xs, Y, Ys, p_out(X, Y)) → U2(X, Xs, Y, Ys, map_in(Xs, Ys))
U2(X, Xs, Y, Ys, map_out(Xs, Ys)) → map_out(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in(x1, x2)  =  map_in(x1)
[]  =  []
map_out(x1, x2)  =  map_out(x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, 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:

MAP_IN(.(X, Xs), .(Y, Ys)) → U11(X, Xs, Y, Ys, p_in(X, Y))
MAP_IN(.(X, Xs), .(Y, Ys)) → P_IN(X, Y)
U11(X, Xs, Y, Ys, p_out(X, Y)) → U21(X, Xs, Y, Ys, map_in(Xs, Ys))
U11(X, Xs, Y, Ys, p_out(X, Y)) → MAP_IN(Xs, Ys)

The TRS R consists of the following rules:

map_in([], []) → map_out([], [])
map_in(.(X, Xs), .(Y, Ys)) → U1(X, Xs, Y, Ys, p_in(X, Y))
p_in(val_i, val_j) → p_out(val_i, val_j)
U1(X, Xs, Y, Ys, p_out(X, Y)) → U2(X, Xs, Y, Ys, map_in(Xs, Ys))
U2(X, Xs, Y, Ys, map_out(Xs, Ys)) → map_out(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in(x1, x2)  =  map_in(x1)
[]  =  []
map_out(x1, x2)  =  map_out(x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
P_IN(x1, x2)  =  P_IN(x1)
MAP_IN(x1, x2)  =  MAP_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x2, x5)

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:

MAP_IN(.(X, Xs), .(Y, Ys)) → U11(X, Xs, Y, Ys, p_in(X, Y))
MAP_IN(.(X, Xs), .(Y, Ys)) → P_IN(X, Y)
U11(X, Xs, Y, Ys, p_out(X, Y)) → U21(X, Xs, Y, Ys, map_in(Xs, Ys))
U11(X, Xs, Y, Ys, p_out(X, Y)) → MAP_IN(Xs, Ys)

The TRS R consists of the following rules:

map_in([], []) → map_out([], [])
map_in(.(X, Xs), .(Y, Ys)) → U1(X, Xs, Y, Ys, p_in(X, Y))
p_in(val_i, val_j) → p_out(val_i, val_j)
U1(X, Xs, Y, Ys, p_out(X, Y)) → U2(X, Xs, Y, Ys, map_in(Xs, Ys))
U2(X, Xs, Y, Ys, map_out(Xs, Ys)) → map_out(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in(x1, x2)  =  map_in(x1)
[]  =  []
map_out(x1, x2)  =  map_out(x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
P_IN(x1, x2)  =  P_IN(x1)
MAP_IN(x1, x2)  =  MAP_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x2, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

U11(X, Xs, Y, Ys, p_out(X, Y)) → MAP_IN(Xs, Ys)
MAP_IN(.(X, Xs), .(Y, Ys)) → U11(X, Xs, Y, Ys, p_in(X, Y))

The TRS R consists of the following rules:

map_in([], []) → map_out([], [])
map_in(.(X, Xs), .(Y, Ys)) → U1(X, Xs, Y, Ys, p_in(X, Y))
p_in(val_i, val_j) → p_out(val_i, val_j)
U1(X, Xs, Y, Ys, p_out(X, Y)) → U2(X, Xs, Y, Ys, map_in(Xs, Ys))
U2(X, Xs, Y, Ys, map_out(Xs, Ys)) → map_out(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in(x1, x2)  =  map_in(x1)
[]  =  []
map_out(x1, x2)  =  map_out(x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
MAP_IN(x1, x2)  =  MAP_IN(x1)
U11(x1, x2, x3, x4, x5)  =  U11(x2, x5)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

U11(X, Xs, Y, Ys, p_out(X, Y)) → MAP_IN(Xs, Ys)
MAP_IN(.(X, Xs), .(Y, Ys)) → U11(X, Xs, Y, Ys, p_in(X, Y))

The TRS R consists of the following rules:

p_in(val_i, val_j) → p_out(val_i, val_j)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
p_in(x1, x2)  =  p_in(x1)
val_i  =  val_i
val_j  =  val_j
p_out(x1, x2)  =  p_out(x2)
MAP_IN(x1, x2)  =  MAP_IN(x1)
U11(x1, x2, x3, x4, x5)  =  U11(x2, 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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

MAP_IN(.(X, Xs)) → U11(Xs, p_in(X))
U11(Xs, p_out(Y)) → MAP_IN(Xs)

The TRS R consists of the following rules:

p_in(val_i) → p_out(val_j)

The set Q consists of the following terms:

p_in(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: