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



↳ PROLOG
  ↳ PrologToPiTRSProof

minimum2(tree3(X, void0, underscore), X).
minimum2(tree3(underscore1, Left, underscore2), X) :- minimum2(Left, X).


With regard to the inferred argument filtering the predicates were used in the following modes:
minimum2: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


minimum_2_in_ga2(tree_33(X, void_0, underscore), X) -> minimum_2_out_ga2(tree_33(X, void_0, underscore), X)
minimum_2_in_ga2(tree_33(underscore1, Left, underscore2), X) -> if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_out_ga2(Left, X)) -> minimum_2_out_ga2(tree_33(underscore1, Left, underscore2), X)

The argument filtering Pi contains the following mapping:
minimum_2_in_ga2(x1, x2)  =  minimum_2_in_ga1(x1)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
void_0  =  void_0
minimum_2_out_ga2(x1, x2)  =  minimum_2_out_ga1(x2)
if_minimum_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_minimum_2_in_1_ga1(x5)

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:

minimum_2_in_ga2(tree_33(X, void_0, underscore), X) -> minimum_2_out_ga2(tree_33(X, void_0, underscore), X)
minimum_2_in_ga2(tree_33(underscore1, Left, underscore2), X) -> if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_out_ga2(Left, X)) -> minimum_2_out_ga2(tree_33(underscore1, Left, underscore2), X)

The argument filtering Pi contains the following mapping:
minimum_2_in_ga2(x1, x2)  =  minimum_2_in_ga1(x1)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
void_0  =  void_0
minimum_2_out_ga2(x1, x2)  =  minimum_2_out_ga1(x2)
if_minimum_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_minimum_2_in_1_ga1(x5)


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

MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> IF_MINIMUM_2_IN_1_GA5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> MINIMUM_2_IN_GA2(Left, X)

The TRS R consists of the following rules:

minimum_2_in_ga2(tree_33(X, void_0, underscore), X) -> minimum_2_out_ga2(tree_33(X, void_0, underscore), X)
minimum_2_in_ga2(tree_33(underscore1, Left, underscore2), X) -> if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_out_ga2(Left, X)) -> minimum_2_out_ga2(tree_33(underscore1, Left, underscore2), X)

The argument filtering Pi contains the following mapping:
minimum_2_in_ga2(x1, x2)  =  minimum_2_in_ga1(x1)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
void_0  =  void_0
minimum_2_out_ga2(x1, x2)  =  minimum_2_out_ga1(x2)
if_minimum_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_minimum_2_in_1_ga1(x5)
MINIMUM_2_IN_GA2(x1, x2)  =  MINIMUM_2_IN_GA1(x1)
IF_MINIMUM_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_MINIMUM_2_IN_1_GA1(x5)

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:

MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> IF_MINIMUM_2_IN_1_GA5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> MINIMUM_2_IN_GA2(Left, X)

The TRS R consists of the following rules:

minimum_2_in_ga2(tree_33(X, void_0, underscore), X) -> minimum_2_out_ga2(tree_33(X, void_0, underscore), X)
minimum_2_in_ga2(tree_33(underscore1, Left, underscore2), X) -> if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_out_ga2(Left, X)) -> minimum_2_out_ga2(tree_33(underscore1, Left, underscore2), X)

The argument filtering Pi contains the following mapping:
minimum_2_in_ga2(x1, x2)  =  minimum_2_in_ga1(x1)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
void_0  =  void_0
minimum_2_out_ga2(x1, x2)  =  minimum_2_out_ga1(x2)
if_minimum_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_minimum_2_in_1_ga1(x5)
MINIMUM_2_IN_GA2(x1, x2)  =  MINIMUM_2_IN_GA1(x1)
IF_MINIMUM_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_MINIMUM_2_IN_1_GA1(x5)

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:

MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> MINIMUM_2_IN_GA2(Left, X)

The TRS R consists of the following rules:

minimum_2_in_ga2(tree_33(X, void_0, underscore), X) -> minimum_2_out_ga2(tree_33(X, void_0, underscore), X)
minimum_2_in_ga2(tree_33(underscore1, Left, underscore2), X) -> if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_in_ga2(Left, X))
if_minimum_2_in_1_ga5(underscore1, Left, underscore2, X, minimum_2_out_ga2(Left, X)) -> minimum_2_out_ga2(tree_33(underscore1, Left, underscore2), X)

The argument filtering Pi contains the following mapping:
minimum_2_in_ga2(x1, x2)  =  minimum_2_in_ga1(x1)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
void_0  =  void_0
minimum_2_out_ga2(x1, x2)  =  minimum_2_out_ga1(x2)
if_minimum_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_minimum_2_in_1_ga1(x5)
MINIMUM_2_IN_GA2(x1, x2)  =  MINIMUM_2_IN_GA1(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:

MINIMUM_2_IN_GA2(tree_33(underscore1, Left, underscore2), X) -> MINIMUM_2_IN_GA2(Left, X)

R is empty.
The argument filtering Pi contains the following mapping:
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
MINIMUM_2_IN_GA2(x1, x2)  =  MINIMUM_2_IN_GA1(x1)

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:

MINIMUM_2_IN_GA1(tree_33(underscore1, Left, underscore2)) -> MINIMUM_2_IN_GA1(Left)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MINIMUM_2_IN_GA1}.
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: