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].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

h_in(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out

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(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out


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(X) → U31(X, f_in(X))
H_IN(X) → F_IN(X)
F_IN(c(s(X), Y)) → U11(X, Y, f_in(c(X, s(Y))))
F_IN(c(s(X), Y)) → F_IN(c(X, s(Y)))
U31(X, f_out(X)) → U41(X, g_in(X))
U31(X, f_out(X)) → G_IN(X)
G_IN(c(X, s(Y))) → U21(X, Y, g_in(c(s(X), Y)))
G_IN(c(X, s(Y))) → G_IN(c(s(X), Y))

The TRS R consists of the following rules:

h_in(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out
G_IN(x1)  =  G_IN(x1)
U41(x1, x2)  =  U41(x2)
U31(x1, x2)  =  U31(x1, x2)
U21(x1, x2, x3)  =  U21(x3)
H_IN(x1)  =  H_IN(x1)
U11(x1, x2, x3)  =  U11(x3)
F_IN(x1)  =  F_IN(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(X) → U31(X, f_in(X))
H_IN(X) → F_IN(X)
F_IN(c(s(X), Y)) → U11(X, Y, f_in(c(X, s(Y))))
F_IN(c(s(X), Y)) → F_IN(c(X, s(Y)))
U31(X, f_out(X)) → U41(X, g_in(X))
U31(X, f_out(X)) → G_IN(X)
G_IN(c(X, s(Y))) → U21(X, Y, g_in(c(s(X), Y)))
G_IN(c(X, s(Y))) → G_IN(c(s(X), Y))

The TRS R consists of the following rules:

h_in(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out
G_IN(x1)  =  G_IN(x1)
U41(x1, x2)  =  U41(x2)
U31(x1, x2)  =  U31(x1, x2)
U21(x1, x2, x3)  =  U21(x3)
H_IN(x1)  =  H_IN(x1)
U11(x1, x2, x3)  =  U11(x3)
F_IN(x1)  =  F_IN(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(c(X, s(Y))) → G_IN(c(s(X), Y))

The TRS R consists of the following rules:

h_in(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out
G_IN(x1)  =  G_IN(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(c(X, s(Y))) → G_IN(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(c(X, s(Y))) → G_IN(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(c(X, s(Y))) → G_IN(c(s(X), Y))


Used ordering: POLO with Polynomial interpretation [25]:

POL(G_IN(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(c(s(X), Y)) → F_IN(c(X, s(Y)))

The TRS R consists of the following rules:

h_in(X) → U3(X, f_in(X))
f_in(c(s(X), Y)) → U1(X, Y, f_in(c(X, s(Y))))
U1(X, Y, f_out(c(X, s(Y)))) → f_out(c(s(X), Y))
U3(X, f_out(X)) → U4(X, g_in(X))
g_in(c(X, s(Y))) → U2(X, Y, g_in(c(s(X), Y)))
U2(X, Y, g_out(c(s(X), Y))) → g_out(c(X, s(Y)))
U4(X, g_out(X)) → h_out(X)

The argument filtering Pi contains the following mapping:
h_in(x1)  =  h_in(x1)
U3(x1, x2)  =  U3(x1, x2)
f_in(x1)  =  f_in(x1)
c(x1, x2)  =  c(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3)  =  U1(x3)
f_out(x1)  =  f_out
U4(x1, x2)  =  U4(x2)
g_in(x1)  =  g_in(x1)
U2(x1, x2, x3)  =  U2(x3)
g_out(x1)  =  g_out
h_out(x1)  =  h_out
F_IN(x1)  =  F_IN(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(c(s(X), Y)) → F_IN(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(c(s(X), Y)) → F_IN(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(c(s(X), Y)) → F_IN(c(X, s(Y)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(F_IN(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.