Left Termination of the query pattern
map(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).

Query: map(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: map(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: map(T1, T2), SCOPE: 1, CLAUSE: 0 | map(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(!_1, eq(T5, [])) | map([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: map(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX5 is free\nmap(T1, T2) !=? map([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: eq(T5, [])\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: eq(T5, []), SCOPE: 2, CLAUSE: 7\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(head(T12, X15), ','(tail(T12, X16), ','(p(X15, T15), map(X16, T16))))\n\nT12 is ground\nX5 is free\nX15 is free\nX16 is free\nmap(T12, T2) !=? map([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: ','(head(T12, X15), ','(tail(T12, X16), ','(p(X15, T15), map(X16, T16)))), SCOPE: 3, CLAUSE: 2 | ','(head(T12, X15), ','(tail(T12, X16), ','(p(X15, T15), map(X16, T16)))), SCOPE: 3, CLAUSE: 3\n\nT12 is ground\nX5 is free\nX15 is free\nX16 is free\nmap(T12, T2) !=? map([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: ','(head(T12, X15), ','(tail(T12, X16), ','(p(X15, T15), map(X16, T16)))), SCOPE: 3, CLAUSE: 3\n\nT12 is ground\nX5 is free\nX15 is free\nX16 is free\nmap(T12, T2) !=? map([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: ','(tail(.(T21, T22), X16), ','(p(T21, T15), map(X16, T16)))\n\nT21 is ground\nT22 is ground\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: ','(tail(.(T21, T22), X16), ','(p(T21, T15), map(X16, T16))), SCOPE: 4, CLAUSE: 4 | ','(tail(.(T21, T22), X16), ','(p(T21, T15), map(X16, T16))), SCOPE: 4, CLAUSE: 5\n\nT21 is ground\nT22 is ground\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: ','(tail(.(T21, T22), X16), ','(p(T21, T15), map(X16, T16))), SCOPE: 4, CLAUSE: 5\n\nT21 is ground\nT22 is ground\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: ','(p(T29, T15), map(T30, T16))\n\nT29 is ground\nT30 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="51: ','(p(T29, T15), map(T30, T16)), SCOPE: 5, CLAUSE: 6\n\nT29 is ground\nT30 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
56 [label="56: map(T30, T40)\n\nT30 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
57 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 57 [arrowhead = none ];
57 -> 5;

58 [label="EVAL with clause\nmap([], X5) :- ','(!_1, eq(X5, [])).\nand substitutionT1 -> [],\nT2 -> T5,\nX5 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 58 [arrowhead = none ];
58 -> 8;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 59 [arrowhead = none ];
59 -> 11;

60 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 60 [arrowhead = none ];
60 -> 14;

61 [label="EVAL with clause\nmap(X12, .(X13, X14)) :- ','(head(X12, X15), ','(tail(X12, X16), ','(p(X15, X13), map(X16, X14)))).\nand substitutionT1 -> T12,\nX12 -> T12,\nX13 -> T15,\nX14 -> T16,\nT2 -> .(T15, T16),\nT13 -> T15,\nT14 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 61 [arrowhead = none ];
61 -> 28;

62 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 62 [arrowhead = none ];
62 -> 30;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 63 [arrowhead = none ];
63 -> 16;

64 [label="EVAL with clause\neq(X8, X8).\nand substitutionT5 -> [],\nX8 -> [],\nT8 -> []", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 64 [arrowhead = none ];
64 -> 19;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 65 [arrowhead = none ];
65 -> 22;

66 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 66 [arrowhead = none ];
66 -> 23;

67 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 67 [arrowhead = none ];
67 -> 33;

68 [label="BACKTRACK\nfor clause: head([], X1)\nwith clash: (map(T12, T2), map([], X5))", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 68 [arrowhead = none ];
68 -> 35;

69 [label="EVAL with clause\nhead(.(X23, X24), X23).\nand substitutionX23 -> T21,\nX24 -> T22,\nT12 -> .(T21, T22),\nX15 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 69 [arrowhead = none ];
69 -> 40;

70 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 70 [arrowhead = none ];
70 -> 41;

71 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
40 -> 71 [arrowhead = none ];
71 -> 42;

72 [label="BACKTRACK\nfor clause: tail([], [])because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
42 -> 72 [arrowhead = none ];
72 -> 44;

73 [label="ONLY EVAL with clause\ntail(.(X31, X32), X32).\nand substitutionT21 -> T29,\nX31 -> T29,\nT22 -> T30,\nX32 -> T30,\nX16 -> T30", fontsize=14, style = filled, fillcolor = lightblue];
44 -> 73 [arrowhead = none ];
73 -> 50;

74 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
50 -> 74 [arrowhead = none ];
74 -> 51;

75 [label="ONLY EVAL with clause\np(X39, X40).\nand substitutionT29 -> T38,\nX39 -> T38,\nT15 -> T39,\nX40 -> T39,\nT16 -> T40", fontsize=14, style = filled, fillcolor = lightblue];
51 -> 75 [arrowhead = none ];
75 -> 56;

76 [label="INSTANCE with matching:\nT1 -> T30\nT2 -> T40", fontsize=14, style = filled, fillcolor = lightgrey];
56 -> 76 [arrowhead = none , style=dashed];
76 -> 1[style=dashed];

}

(2) Obligation:

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

mapA_in_ga([], []) → mapA_out_ga([], [])
mapA_in_ga(.(T38, T30), .(T39, T40)) → U1_ga(T38, T30, T39, T40, mapA_in_ga(T30, T40))
U1_ga(T38, T30, T39, T40, mapA_out_ga(T30, T40)) → mapA_out_ga(.(T38, T30), .(T39, T40))

The argument filtering Pi contains the following mapping:
mapA_in_ga(x1, x2)  =  mapA_in_ga(x1)
[]  =  []
mapA_out_ga(x1, x2)  =  mapA_out_ga(x1, x2)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

MAPA_IN_GA(.(T38, T30), .(T39, T40)) → U1_GA(T38, T30, T39, T40, mapA_in_ga(T30, T40))
MAPA_IN_GA(.(T38, T30), .(T39, T40)) → MAPA_IN_GA(T30, T40)

The TRS R consists of the following rules:

mapA_in_ga([], []) → mapA_out_ga([], [])
mapA_in_ga(.(T38, T30), .(T39, T40)) → U1_ga(T38, T30, T39, T40, mapA_in_ga(T30, T40))
U1_ga(T38, T30, T39, T40, mapA_out_ga(T30, T40)) → mapA_out_ga(.(T38, T30), .(T39, T40))

The argument filtering Pi contains the following mapping:
mapA_in_ga(x1, x2)  =  mapA_in_ga(x1)
[]  =  []
mapA_out_ga(x1, x2)  =  mapA_out_ga(x1, x2)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
MAPA_IN_GA(x1, x2)  =  MAPA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(4) Obligation:

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

MAPA_IN_GA(.(T38, T30), .(T39, T40)) → U1_GA(T38, T30, T39, T40, mapA_in_ga(T30, T40))
MAPA_IN_GA(.(T38, T30), .(T39, T40)) → MAPA_IN_GA(T30, T40)

The TRS R consists of the following rules:

mapA_in_ga([], []) → mapA_out_ga([], [])
mapA_in_ga(.(T38, T30), .(T39, T40)) → U1_ga(T38, T30, T39, T40, mapA_in_ga(T30, T40))
U1_ga(T38, T30, T39, T40, mapA_out_ga(T30, T40)) → mapA_out_ga(.(T38, T30), .(T39, T40))

The argument filtering Pi contains the following mapping:
mapA_in_ga(x1, x2)  =  mapA_in_ga(x1)
[]  =  []
mapA_out_ga(x1, x2)  =  mapA_out_ga(x1, x2)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
MAPA_IN_GA(x1, x2)  =  MAPA_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:

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

The TRS R consists of the following rules:

mapA_in_ga([], []) → mapA_out_ga([], [])
mapA_in_ga(.(T38, T30), .(T39, T40)) → U1_ga(T38, T30, T39, T40, mapA_in_ga(T30, T40))
U1_ga(T38, T30, T39, T40, mapA_out_ga(T30, T40)) → mapA_out_ga(.(T38, T30), .(T39, T40))

The argument filtering Pi contains the following mapping:
mapA_in_ga(x1, x2)  =  mapA_in_ga(x1)
[]  =  []
mapA_out_ga(x1, x2)  =  mapA_out_ga(x1, x2)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
MAPA_IN_GA(x1, x2)  =  MAPA_IN_GA(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

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

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

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

MAPA_IN_GA(.(T30)) → MAPA_IN_GA(T30)

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

(11) 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:

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

(12) YES