Left Termination of the query pattern h_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

f(c(s(X), Y)) :- f(c(X, s(Y))).
g(c(X, s(Y))) :- g(c(s(X), Y)).
h(X) :- ','(f(X), g(X)).

Queries:

h(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
h_in: (b)
f_in: (b)
g_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g

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:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

H_IN_G(X) → U3_G(X, f_in_g(X))
H_IN_G(X) → F_IN_G(X)
F_IN_G(c(s(X), Y)) → U1_G(X, Y, f_in_g(c(X, s(Y))))
F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))
U3_G(X, f_out_g(X)) → U4_G(X, g_in_g(X))
U3_G(X, f_out_g(X)) → G_IN_G(X)
G_IN_G(c(X, s(Y))) → U2_G(X, Y, g_in_g(c(s(X), Y)))
G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))

The TRS R consists of the following rules:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g
U2_G(x1, x2, x3)  =  U2_G(x3)
F_IN_G(x1)  =  F_IN_G(x1)
U4_G(x1, x2)  =  U4_G(x2)
H_IN_G(x1)  =  H_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x3)
U3_G(x1, x2)  =  U3_G(x1, x2)
G_IN_G(x1)  =  G_IN_G(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:

H_IN_G(X) → U3_G(X, f_in_g(X))
H_IN_G(X) → F_IN_G(X)
F_IN_G(c(s(X), Y)) → U1_G(X, Y, f_in_g(c(X, s(Y))))
F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))
U3_G(X, f_out_g(X)) → U4_G(X, g_in_g(X))
U3_G(X, f_out_g(X)) → G_IN_G(X)
G_IN_G(c(X, s(Y))) → U2_G(X, Y, g_in_g(c(s(X), Y)))
G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))

The TRS R consists of the following rules:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g
U2_G(x1, x2, x3)  =  U2_G(x3)
F_IN_G(x1)  =  F_IN_G(x1)
U4_G(x1, x2)  =  U4_G(x2)
H_IN_G(x1)  =  H_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x3)
U3_G(x1, x2)  =  U3_G(x1, x2)
G_IN_G(x1)  =  G_IN_G(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.

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

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

G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))

The TRS R consists of the following rules:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g
G_IN_G(x1)  =  G_IN_G(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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))

R is empty.
Pi is empty.
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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
              ↳ PiDP

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

G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

G_IN_G(c(X, s(Y))) → G_IN_G(c(s(X), Y))


Used ordering: POLO with Polynomial interpretation [25]:

POL(G_IN_G(x1)) = 2·x1   
POL(c(x1, x2)) = x1 + 2·x2   
POL(s(x1)) = 1 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ PisEmptyProof
              ↳ PiDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))

The TRS R consists of the following rules:

h_in_g(X) → U3_g(X, f_in_g(X))
f_in_g(c(s(X), Y)) → U1_g(X, Y, f_in_g(c(X, s(Y))))
U1_g(X, Y, f_out_g(c(X, s(Y)))) → f_out_g(c(s(X), Y))
U3_g(X, f_out_g(X)) → U4_g(X, g_in_g(X))
g_in_g(c(X, s(Y))) → U2_g(X, Y, g_in_g(c(s(X), Y)))
U2_g(X, Y, g_out_g(c(s(X), Y))) → g_out_g(c(X, s(Y)))
U4_g(X, g_out_g(X)) → h_out_g(X)

The argument filtering Pi contains the following mapping:
h_in_g(x1)  =  h_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
f_in_g(x1)  =  f_in_g(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
f_out_g(x1)  =  f_out_g
U4_g(x1, x2)  =  U4_g(x2)
g_in_g(x1)  =  g_in_g(x1)
U2_g(x1, x2, x3)  =  U2_g(x3)
g_out_g(x1)  =  g_out_g
h_out_g(x1)  =  h_out_g
F_IN_G(x1)  =  F_IN_G(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))

R is empty.
Pi is empty.
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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F_IN_G(c(s(X), Y)) → F_IN_G(c(X, s(Y)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(F_IN_G(x1)) = 2·x1   
POL(c(x1, x2)) = 2·x1 + x2   
POL(s(x1)) = 1 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.