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

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_ga(X, Y))
p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)
U1_ga(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x3, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)

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_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_ga(X, Y))
p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)
U1_ga(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x3, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)


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_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_ga(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_GA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_ga(X, Y))
p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)
U1_ga(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x3, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x3, x5)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)

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_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_ga(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_GA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_ga(X, Y))
p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)
U1_ga(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x3, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x3, x5)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)

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:

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_ga(X, Y))
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_ga(X, Y))
p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)
U1_ga(X, Xs, Y, Ys, p_out_ga(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x3, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)

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:

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_ga(X, Y))
U1_GA(X, Xs, Y, Ys, p_out_ga(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

p_in_ga(val_i, val_j) → p_out_ga(val_i, val_j)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
val_i  =  val_i
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)

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:

U1_GA(Xs, p_out_ga(Y)) → MAP_IN_GA(Xs)
MAP_IN_GA(.(X, Xs)) → U1_GA(Xs, p_in_ga(X))

The TRS R consists of the following rules:

p_in_ga(val_i) → p_out_ga(val_j)

The set Q consists of the following terms:

p_in_ga(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: