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

(0) Obligation:

Clauses:

p(X, X).
p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y))).

Query: p(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: p(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: p(T1, T2), SCOPE: 1, CLAUSE: 0 | p(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | p(T4, T2), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: p(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX2 is free\np(T1, T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: p(T4, T2), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(p(f(T7), f(X7)), p(X7, g(T9)))\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 0 | ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 0\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: p(T22, g(T9))\n\nT22 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: ','(p(f(T30), f(X25)), p(X25, g(T32)))\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 0 | ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 1\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 0\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="51: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 1\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
54 [label="54: p(T45, g(T32))\n\nT45 is ground\nX2 is free\np(f(T45), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
55 [label="55: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
56 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 56 [arrowhead = none ];
56 -> 4;

57 [label="EVAL with clause\np(X2, X2).\nand substitutionT1 -> T4,\nX2 -> T4,\nT2 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 57 [arrowhead = none ];
57 -> 9;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 58 [arrowhead = none ];
58 -> 14;

59 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 59 [arrowhead = none ];
59 -> 17;

60 [label="EVAL with clause\np(f(X23), g(X24)) :- ','(p(f(X23), f(X25)), p(X25, g(X24))).\nand substitutionX23 -> T30,\nT1 -> f(T30),\nX24 -> T32,\nT2 -> g(T32),\nT31 -> T32", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 60 [arrowhead = none ];
60 -> 38;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 61 [arrowhead = none ];
61 -> 43;

62 [label="EVAL with clause\np(f(X5), g(X6)) :- ','(p(f(X5), f(X7)), p(X7, g(X6))).\nand substitutionX5 -> T7,\nT4 -> f(T7),\nX6 -> T9,\nT2 -> g(T9),\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 62 [arrowhead = none ];
62 -> 19;

63 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 63 [arrowhead = none ];
63 -> 21;

64 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 64 [arrowhead = none ];
64 -> 24;

65 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
24 -> 65 [arrowhead = none ];
65 -> 27;

66 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
24 -> 66 [arrowhead = none ];
66 -> 29;

67 [label="ONLY EVAL with clause\np(X16, X16).\nand substitutionT7 -> T22,\nX16 -> f(T22),\nX7 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 67 [arrowhead = none ];
67 -> 31;

68 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 68 [arrowhead = none ];
68 -> 35;

69 [label="INSTANCE with matching:\nT1 -> T22\nT2 -> g(T9)", fontsize=14, style = filled, fillcolor = lightgrey];
31 -> 69 [arrowhead = none , style=dashed];
69 -> 3[style=dashed];

70 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
38 -> 70 [arrowhead = none ];
70 -> 46;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
46 -> 71 [arrowhead = none ];
71 -> 50;

72 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
46 -> 72 [arrowhead = none ];
72 -> 51;

73 [label="ONLY EVAL with clause\np(X34, X34).\nand substitutionT30 -> T45,\nX34 -> f(T45),\nX25 -> T45", fontsize=14, style = filled, fillcolor = lightblue];
50 -> 73 [arrowhead = none ];
73 -> 54;

74 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
51 -> 74 [arrowhead = none ];
74 -> 55;

75 [label="INSTANCE with matching:\nT1 -> T45\nT2 -> g(T32)", fontsize=14, style = filled, fillcolor = lightgrey];
54 -> 75 [arrowhead = none , style=dashed];
75 -> 3[style=dashed];

}

(2) Obligation:

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

pA_in_ga(T4, T4) → pA_out_ga(T4, T4)
pA_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(T4, T4) → pA_out_gg(T4, T4)
pA_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, pA_in_gg(T45, g(T32)))
U2_gg(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_gg(f(T45), g(T32))
U1_gg(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_gg(f(T22), g(T9))
U1_ga(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_ga(f(T22), g(T9))
pA_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, pA_in_gg(T45, g(T32)))
U2_ga(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_ga(f(T45), g(T32))

The argument filtering Pi contains the following mapping:
pA_in_ga(x1, x2)  =  pA_in_ga(x1)
pA_out_ga(x1, x2)  =  pA_out_ga(x1, x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
g(x1)  =  g
pA_out_gg(x1, x2)  =  pA_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)

(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:

PA_IN_GA(f(T22), g(T9)) → U1_GA(T22, T9, pA_in_gg(T22, g(T9)))
PA_IN_GA(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))
PA_IN_GG(f(T22), g(T9)) → U1_GG(T22, T9, pA_in_gg(T22, g(T9)))
PA_IN_GG(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))
PA_IN_GG(f(T45), g(T32)) → U2_GG(T45, T32, pA_in_gg(T45, g(T32)))
PA_IN_GA(f(T45), g(T32)) → U2_GA(T45, T32, pA_in_gg(T45, g(T32)))

The TRS R consists of the following rules:

pA_in_ga(T4, T4) → pA_out_ga(T4, T4)
pA_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(T4, T4) → pA_out_gg(T4, T4)
pA_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, pA_in_gg(T45, g(T32)))
U2_gg(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_gg(f(T45), g(T32))
U1_gg(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_gg(f(T22), g(T9))
U1_ga(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_ga(f(T22), g(T9))
pA_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, pA_in_gg(T45, g(T32)))
U2_ga(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_ga(f(T45), g(T32))

The argument filtering Pi contains the following mapping:
pA_in_ga(x1, x2)  =  pA_in_ga(x1)
pA_out_ga(x1, x2)  =  pA_out_ga(x1, x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
g(x1)  =  g
pA_out_gg(x1, x2)  =  pA_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
PA_IN_GA(x1, x2)  =  PA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
PA_IN_GG(x1, x2)  =  PA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x3)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(4) Obligation:

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

PA_IN_GA(f(T22), g(T9)) → U1_GA(T22, T9, pA_in_gg(T22, g(T9)))
PA_IN_GA(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))
PA_IN_GG(f(T22), g(T9)) → U1_GG(T22, T9, pA_in_gg(T22, g(T9)))
PA_IN_GG(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))
PA_IN_GG(f(T45), g(T32)) → U2_GG(T45, T32, pA_in_gg(T45, g(T32)))
PA_IN_GA(f(T45), g(T32)) → U2_GA(T45, T32, pA_in_gg(T45, g(T32)))

The TRS R consists of the following rules:

pA_in_ga(T4, T4) → pA_out_ga(T4, T4)
pA_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(T4, T4) → pA_out_gg(T4, T4)
pA_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, pA_in_gg(T45, g(T32)))
U2_gg(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_gg(f(T45), g(T32))
U1_gg(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_gg(f(T22), g(T9))
U1_ga(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_ga(f(T22), g(T9))
pA_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, pA_in_gg(T45, g(T32)))
U2_ga(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_ga(f(T45), g(T32))

The argument filtering Pi contains the following mapping:
pA_in_ga(x1, x2)  =  pA_in_ga(x1)
pA_out_ga(x1, x2)  =  pA_out_ga(x1, x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
g(x1)  =  g
pA_out_gg(x1, x2)  =  pA_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
PA_IN_GA(x1, x2)  =  PA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
PA_IN_GG(x1, x2)  =  PA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x3)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(6) Obligation:

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

PA_IN_GG(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))

The TRS R consists of the following rules:

pA_in_ga(T4, T4) → pA_out_ga(T4, T4)
pA_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(T4, T4) → pA_out_gg(T4, T4)
pA_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, pA_in_gg(T22, g(T9)))
pA_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, pA_in_gg(T45, g(T32)))
U2_gg(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_gg(f(T45), g(T32))
U1_gg(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_gg(f(T22), g(T9))
U1_ga(T22, T9, pA_out_gg(T22, g(T9))) → pA_out_ga(f(T22), g(T9))
pA_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, pA_in_gg(T45, g(T32)))
U2_ga(T45, T32, pA_out_gg(T45, g(T32))) → pA_out_ga(f(T45), g(T32))

The argument filtering Pi contains the following mapping:
pA_in_ga(x1, x2)  =  pA_in_ga(x1)
pA_out_ga(x1, x2)  =  pA_out_ga(x1, x2)
f(x1)  =  f(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
g(x1)  =  g
pA_out_gg(x1, x2)  =  pA_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
PA_IN_GG(x1, x2)  =  PA_IN_GG(x1, x2)

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:

PA_IN_GG(f(T22), g(T9)) → PA_IN_GG(T22, g(T9))

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

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:

PA_IN_GG(f(T22), g) → PA_IN_GG(T22, g)

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:

• PA_IN_GG(f(T22), g) → PA_IN_GG(T22, g)
  The graph contains the following edges 1 > 1, 2 >= 2

(12) YES