Left Termination of the query pattern less(b,f) w.r.t. the given Prolog program could not be shown:



PROLOG
  ↳ PrologToPiTRSProof
  ↳ 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: (b,f) (f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga1(x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga1(x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)


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

LESS_2_IN_GA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_GA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga1(x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_GA2(x1, x2)  =  LESS_2_IN_GA1(x1)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA
IF_LESS_2_IN_1_AA3(x1, x2, x3)  =  IF_LESS_2_IN_1_AA1(x3)
IF_LESS_2_IN_1_GA3(x1, x2, x3)  =  IF_LESS_2_IN_1_GA1(x3)

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

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

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

LESS_2_IN_GA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_GA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga1(x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_GA2(x1, x2)  =  LESS_2_IN_GA1(x1)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA
IF_LESS_2_IN_1_AA3(x1, x2, x3)  =  IF_LESS_2_IN_1_AA1(x3)
IF_LESS_2_IN_1_GA3(x1, x2, x3)  =  IF_LESS_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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

LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga1(x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA

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
  ↳ PrologToPiTRSProof

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

LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_1
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA

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
  ↳ PrologToPiTRSProof

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

LESS_2_IN_AA -> LESS_2_IN_AA

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

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga2(x1, x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga2(x1, x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)


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

LESS_2_IN_GA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_GA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga2(x1, x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_GA2(x1, x2)  =  LESS_2_IN_GA1(x1)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA
IF_LESS_2_IN_1_AA3(x1, x2, x3)  =  IF_LESS_2_IN_1_AA1(x3)
IF_LESS_2_IN_1_GA3(x1, x2, x3)  =  IF_LESS_2_IN_1_GA1(x3)

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

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

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

LESS_2_IN_GA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_GA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_GA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> IF_LESS_2_IN_1_AA3(X, Y, less_2_in_aa2(X, Y))
LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga2(x1, x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_GA2(x1, x2)  =  LESS_2_IN_GA1(x1)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA
IF_LESS_2_IN_1_AA3(x1, x2, x3)  =  IF_LESS_2_IN_1_AA1(x3)
IF_LESS_2_IN_1_GA3(x1, x2, x3)  =  IF_LESS_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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

LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

The TRS R consists of the following rules:

less_2_in_ga2(0_0, s_11(underscore)) -> less_2_out_ga2(0_0, s_11(underscore))
less_2_in_ga2(s_11(X), s_11(Y)) -> if_less_2_in_1_ga3(X, Y, less_2_in_aa2(X, Y))
less_2_in_aa2(0_0, s_11(underscore)) -> less_2_out_aa2(0_0, s_11(underscore))
less_2_in_aa2(s_11(X), s_11(Y)) -> if_less_2_in_1_aa3(X, Y, less_2_in_aa2(X, Y))
if_less_2_in_1_aa3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_aa2(s_11(X), s_11(Y))
if_less_2_in_1_ga3(X, Y, less_2_out_aa2(X, Y)) -> less_2_out_ga2(s_11(X), s_11(Y))

The argument filtering Pi contains the following mapping:
less_2_in_ga2(x1, x2)  =  less_2_in_ga1(x1)
0_0  =  0_0
s_11(x1)  =  s_1
less_2_out_ga2(x1, x2)  =  less_2_out_ga2(x1, x2)
if_less_2_in_1_ga3(x1, x2, x3)  =  if_less_2_in_1_ga1(x3)
less_2_in_aa2(x1, x2)  =  less_2_in_aa
less_2_out_aa2(x1, x2)  =  less_2_out_aa2(x1, x2)
if_less_2_in_1_aa3(x1, x2, x3)  =  if_less_2_in_1_aa1(x3)
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA

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

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

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

LESS_2_IN_AA2(s_11(X), s_11(Y)) -> LESS_2_IN_AA2(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_1
LESS_2_IN_AA2(x1, x2)  =  LESS_2_IN_AA

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

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

LESS_2_IN_AA -> LESS_2_IN_AA

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_AA}.