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



↳ PROLOG
  ↳ PrologToPiTRSProof

even1(00).
even1(s1(X)) :- odd1(X).
odd1(s1(X)) :- even1(X).


With regard to the inferred argument filtering the predicates were used in the following modes:
even1: (b)
odd1: (b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


even_1_in_g1(0_0) -> even_1_out_g1(0_0)
even_1_in_g1(s_11(X)) -> if_even_1_in_1_g2(X, odd_1_in_g1(X))
odd_1_in_g1(s_11(X)) -> if_odd_1_in_1_g2(X, even_1_in_g1(X))
if_odd_1_in_1_g2(X, even_1_out_g1(X)) -> odd_1_out_g1(s_11(X))
if_even_1_in_1_g2(X, odd_1_out_g1(X)) -> even_1_out_g1(s_11(X))

The argument filtering Pi contains the following mapping:
even_1_in_g1(x1)  =  even_1_in_g1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
even_1_out_g1(x1)  =  even_1_out_g
if_even_1_in_1_g2(x1, x2)  =  if_even_1_in_1_g1(x2)
odd_1_in_g1(x1)  =  odd_1_in_g1(x1)
if_odd_1_in_1_g2(x1, x2)  =  if_odd_1_in_1_g1(x2)
odd_1_out_g1(x1)  =  odd_1_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:

even_1_in_g1(0_0) -> even_1_out_g1(0_0)
even_1_in_g1(s_11(X)) -> if_even_1_in_1_g2(X, odd_1_in_g1(X))
odd_1_in_g1(s_11(X)) -> if_odd_1_in_1_g2(X, even_1_in_g1(X))
if_odd_1_in_1_g2(X, even_1_out_g1(X)) -> odd_1_out_g1(s_11(X))
if_even_1_in_1_g2(X, odd_1_out_g1(X)) -> even_1_out_g1(s_11(X))

The argument filtering Pi contains the following mapping:
even_1_in_g1(x1)  =  even_1_in_g1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
even_1_out_g1(x1)  =  even_1_out_g
if_even_1_in_1_g2(x1, x2)  =  if_even_1_in_1_g1(x2)
odd_1_in_g1(x1)  =  odd_1_in_g1(x1)
if_odd_1_in_1_g2(x1, x2)  =  if_odd_1_in_1_g1(x2)
odd_1_out_g1(x1)  =  odd_1_out_g


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

EVEN_1_IN_G1(s_11(X)) -> IF_EVEN_1_IN_1_G2(X, odd_1_in_g1(X))
EVEN_1_IN_G1(s_11(X)) -> ODD_1_IN_G1(X)
ODD_1_IN_G1(s_11(X)) -> IF_ODD_1_IN_1_G2(X, even_1_in_g1(X))
ODD_1_IN_G1(s_11(X)) -> EVEN_1_IN_G1(X)

The TRS R consists of the following rules:

even_1_in_g1(0_0) -> even_1_out_g1(0_0)
even_1_in_g1(s_11(X)) -> if_even_1_in_1_g2(X, odd_1_in_g1(X))
odd_1_in_g1(s_11(X)) -> if_odd_1_in_1_g2(X, even_1_in_g1(X))
if_odd_1_in_1_g2(X, even_1_out_g1(X)) -> odd_1_out_g1(s_11(X))
if_even_1_in_1_g2(X, odd_1_out_g1(X)) -> even_1_out_g1(s_11(X))

The argument filtering Pi contains the following mapping:
even_1_in_g1(x1)  =  even_1_in_g1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
even_1_out_g1(x1)  =  even_1_out_g
if_even_1_in_1_g2(x1, x2)  =  if_even_1_in_1_g1(x2)
odd_1_in_g1(x1)  =  odd_1_in_g1(x1)
if_odd_1_in_1_g2(x1, x2)  =  if_odd_1_in_1_g1(x2)
odd_1_out_g1(x1)  =  odd_1_out_g
IF_ODD_1_IN_1_G2(x1, x2)  =  IF_ODD_1_IN_1_G1(x2)
ODD_1_IN_G1(x1)  =  ODD_1_IN_G1(x1)
IF_EVEN_1_IN_1_G2(x1, x2)  =  IF_EVEN_1_IN_1_G1(x2)
EVEN_1_IN_G1(x1)  =  EVEN_1_IN_G1(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:

EVEN_1_IN_G1(s_11(X)) -> IF_EVEN_1_IN_1_G2(X, odd_1_in_g1(X))
EVEN_1_IN_G1(s_11(X)) -> ODD_1_IN_G1(X)
ODD_1_IN_G1(s_11(X)) -> IF_ODD_1_IN_1_G2(X, even_1_in_g1(X))
ODD_1_IN_G1(s_11(X)) -> EVEN_1_IN_G1(X)

The TRS R consists of the following rules:

even_1_in_g1(0_0) -> even_1_out_g1(0_0)
even_1_in_g1(s_11(X)) -> if_even_1_in_1_g2(X, odd_1_in_g1(X))
odd_1_in_g1(s_11(X)) -> if_odd_1_in_1_g2(X, even_1_in_g1(X))
if_odd_1_in_1_g2(X, even_1_out_g1(X)) -> odd_1_out_g1(s_11(X))
if_even_1_in_1_g2(X, odd_1_out_g1(X)) -> even_1_out_g1(s_11(X))

The argument filtering Pi contains the following mapping:
even_1_in_g1(x1)  =  even_1_in_g1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
even_1_out_g1(x1)  =  even_1_out_g
if_even_1_in_1_g2(x1, x2)  =  if_even_1_in_1_g1(x2)
odd_1_in_g1(x1)  =  odd_1_in_g1(x1)
if_odd_1_in_1_g2(x1, x2)  =  if_odd_1_in_1_g1(x2)
odd_1_out_g1(x1)  =  odd_1_out_g
IF_ODD_1_IN_1_G2(x1, x2)  =  IF_ODD_1_IN_1_G1(x2)
ODD_1_IN_G1(x1)  =  ODD_1_IN_G1(x1)
IF_EVEN_1_IN_1_G2(x1, x2)  =  IF_EVEN_1_IN_1_G1(x2)
EVEN_1_IN_G1(x1)  =  EVEN_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

ODD_1_IN_G1(s_11(X)) -> EVEN_1_IN_G1(X)
EVEN_1_IN_G1(s_11(X)) -> ODD_1_IN_G1(X)

The TRS R consists of the following rules:

even_1_in_g1(0_0) -> even_1_out_g1(0_0)
even_1_in_g1(s_11(X)) -> if_even_1_in_1_g2(X, odd_1_in_g1(X))
odd_1_in_g1(s_11(X)) -> if_odd_1_in_1_g2(X, even_1_in_g1(X))
if_odd_1_in_1_g2(X, even_1_out_g1(X)) -> odd_1_out_g1(s_11(X))
if_even_1_in_1_g2(X, odd_1_out_g1(X)) -> even_1_out_g1(s_11(X))

The argument filtering Pi contains the following mapping:
even_1_in_g1(x1)  =  even_1_in_g1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
even_1_out_g1(x1)  =  even_1_out_g
if_even_1_in_1_g2(x1, x2)  =  if_even_1_in_1_g1(x2)
odd_1_in_g1(x1)  =  odd_1_in_g1(x1)
if_odd_1_in_1_g2(x1, x2)  =  if_odd_1_in_1_g1(x2)
odd_1_out_g1(x1)  =  odd_1_out_g
ODD_1_IN_G1(x1)  =  ODD_1_IN_G1(x1)
EVEN_1_IN_G1(x1)  =  EVEN_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

ODD_1_IN_G1(s_11(X)) -> EVEN_1_IN_G1(X)
EVEN_1_IN_G1(s_11(X)) -> ODD_1_IN_G1(X)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

ODD_1_IN_G1(s_11(X)) -> EVEN_1_IN_G1(X)
EVEN_1_IN_G1(s_11(X)) -> ODD_1_IN_G1(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {EVEN_1_IN_G1, ODD_1_IN_G1}.
By using the subterm criterion together with the size-change analysis 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: