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



↳ PROLOG
  ↳ PrologToPiTRSProof

less2(00, s1(underscore)).
less2(s1(X), s1(Y)) :- less2(X, Y).


With regard to the inferred argument filtering the predicates were used in the following modes:
less2: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


less_2_in_ag2(0_0, s_11(underscore)) -> less_2_out_ag2(0_0, s_11(underscore))
less_2_in_ag2(s_11(X), s_11(Y)) -> if_less_2_in_1_ag3(X, Y, less_2_in_ag2(X, Y))
if_less_2_in_1_ag3(X, Y, less_2_out_ag2(X, Y)) -> less_2_out_ag2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ag2(x1, x2)  =  less_2_in_ag1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
less_2_out_ag2(x1, x2)  =  less_2_out_ag1(x1)
if_less_2_in_1_ag3(x1, x2, x3)  =  if_less_2_in_1_ag1(x3)

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:

less_2_in_ag2(0_0, s_11(underscore)) -> less_2_out_ag2(0_0, s_11(underscore))
less_2_in_ag2(s_11(X), s_11(Y)) -> if_less_2_in_1_ag3(X, Y, less_2_in_ag2(X, Y))
if_less_2_in_1_ag3(X, Y, less_2_out_ag2(X, Y)) -> less_2_out_ag2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ag2(x1, x2)  =  less_2_in_ag1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
less_2_out_ag2(x1, x2)  =  less_2_out_ag1(x1)
if_less_2_in_1_ag3(x1, x2, x3)  =  if_less_2_in_1_ag1(x3)


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

LESS_2_IN_AG2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AG3(X, Y, less_2_in_ag2(X, Y))
LESS_2_IN_AG2(s_11(X), s_11(Y)) -> LESS_2_IN_AG2(X, Y)

The TRS R consists of the following rules:

less_2_in_ag2(0_0, s_11(underscore)) -> less_2_out_ag2(0_0, s_11(underscore))
less_2_in_ag2(s_11(X), s_11(Y)) -> if_less_2_in_1_ag3(X, Y, less_2_in_ag2(X, Y))
if_less_2_in_1_ag3(X, Y, less_2_out_ag2(X, Y)) -> less_2_out_ag2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ag2(x1, x2)  =  less_2_in_ag1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
less_2_out_ag2(x1, x2)  =  less_2_out_ag1(x1)
if_less_2_in_1_ag3(x1, x2, x3)  =  if_less_2_in_1_ag1(x3)
LESS_2_IN_AG2(x1, x2)  =  LESS_2_IN_AG1(x2)
IF_LESS_2_IN_1_AG3(x1, x2, x3)  =  IF_LESS_2_IN_1_AG1(x3)

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:

LESS_2_IN_AG2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AG3(X, Y, less_2_in_ag2(X, Y))
LESS_2_IN_AG2(s_11(X), s_11(Y)) -> LESS_2_IN_AG2(X, Y)

The TRS R consists of the following rules:

less_2_in_ag2(0_0, s_11(underscore)) -> less_2_out_ag2(0_0, s_11(underscore))
less_2_in_ag2(s_11(X), s_11(Y)) -> if_less_2_in_1_ag3(X, Y, less_2_in_ag2(X, Y))
if_less_2_in_1_ag3(X, Y, less_2_out_ag2(X, Y)) -> less_2_out_ag2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ag2(x1, x2)  =  less_2_in_ag1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
less_2_out_ag2(x1, x2)  =  less_2_out_ag1(x1)
if_less_2_in_1_ag3(x1, x2, x3)  =  if_less_2_in_1_ag1(x3)
LESS_2_IN_AG2(x1, x2)  =  LESS_2_IN_AG1(x2)
IF_LESS_2_IN_1_AG3(x1, x2, x3)  =  IF_LESS_2_IN_1_AG1(x3)

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

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

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

LESS_2_IN_AG2(s_11(X), s_11(Y)) -> LESS_2_IN_AG2(X, Y)

The TRS R consists of the following rules:

less_2_in_ag2(0_0, s_11(underscore)) -> less_2_out_ag2(0_0, s_11(underscore))
less_2_in_ag2(s_11(X), s_11(Y)) -> if_less_2_in_1_ag3(X, Y, less_2_in_ag2(X, Y))
if_less_2_in_1_ag3(X, Y, less_2_out_ag2(X, Y)) -> less_2_out_ag2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ag2(x1, x2)  =  less_2_in_ag1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
less_2_out_ag2(x1, x2)  =  less_2_out_ag1(x1)
if_less_2_in_1_ag3(x1, x2, x3)  =  if_less_2_in_1_ag1(x3)
LESS_2_IN_AG2(x1, x2)  =  LESS_2_IN_AG1(x2)

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:

LESS_2_IN_AG2(s_11(X), s_11(Y)) -> LESS_2_IN_AG2(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
LESS_2_IN_AG2(x1, x2)  =  LESS_2_IN_AG1(x2)

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:

LESS_2_IN_AG1(s_11(Y)) -> LESS_2_IN_AG1(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LESS_2_IN_AG1}.
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: