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

(0) Obligation:

Clauses:

map([], L) :- ','(!, eq(L, [])).
map(X, .(Y, Ys)) :- ','(head(X, H), ','(tail(X, T), ','(p(H, Y), map(T, Ys)))).
head([], X1).
head(.(H, X2), H).
tail([], []).
tail(.(X3, T), T).
p(X, Y).
eq(X, X).

Queries:

map(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

map1(.(T38, T30), .(T39, T40)) :- map1(T30, T40).

Clauses:

mapc1([], []).
mapc1(.(T38, T30), .(T39, T40)) :- mapc1(T30, T40).

Afs:

map1(x1, x2)  =  map1(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
map1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

MAP1_IN_GA(.(T38, T30), .(T39, T40)) → U1_GA(T38, T30, T39, T40, map1_in_ga(T30, T40))
MAP1_IN_GA(.(T38, T30), .(T39, T40)) → MAP1_IN_GA(T30, T40)

R is empty.
The argument filtering Pi contains the following mapping:
map1_in_ga(x1, x2)  =  map1_in_ga(x1)
.(x1, x2)  =  .(x2)
MAP1_IN_GA(x1, x2)  =  MAP1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

MAP1_IN_GA(.(T38, T30), .(T39, T40)) → U1_GA(T38, T30, T39, T40, map1_in_ga(T30, T40))
MAP1_IN_GA(.(T38, T30), .(T39, T40)) → MAP1_IN_GA(T30, T40)

R is empty.
The argument filtering Pi contains the following mapping:
map1_in_ga(x1, x2)  =  map1_in_ga(x1)
.(x1, x2)  =  .(x2)
MAP1_IN_GA(x1, x2)  =  MAP1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

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

MAP1_IN_GA(.(T38, T30), .(T39, T40)) → MAP1_IN_GA(T30, T40)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
MAP1_IN_GA(x1, x2)  =  MAP1_IN_GA(x1)

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

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

MAP1_IN_GA(.(T30)) → MAP1_IN_GA(T30)

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

(9) 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:

• MAP1_IN_GA(.(T30)) → MAP1_IN_GA(T30)
  The graph contains the following edges 1 > 1

(10) YES