Left Termination proof of /tmp/tmpLDjOY3/ordered.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

ordered₁({}₀).
ordered₁(.₂(X, {}₀)).
ordered₁(.₂(X, .₂(Y, Xs))) :- le₂(X, Y), ordered₁(.₂(Y, Xs)).
le₂(s₁(X), s₁(Y)) :- le₂(X, Y).
le₂(0₀, s₁(0₀)).
le₂(0₀, 0₀).

With regard to the inferred argument filtering the predicates were used in the following modes:
ordered₁: (b)
le₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

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:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_in_gg₂(X, Y))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> IF_ORDERED_1_IN_2_G₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

The TRS R consists of the following rules:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g₁(x1) = ordered_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
ordered_1_out_g₁(x1) = ordered_1_out_g
if_ordered_1_in_1_g₄(x1, x2, x3, x4) = if_ordered_1_in_1_g₃(x2, x3, x4)
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
if_ordered_1_in_2_g₄(x1, x2, x3, x4) = if_ordered_1_in_2_g₁(x4)
IF_ORDERED_1_IN_2_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_2_G₁(x4)
IF_ORDERED_1_IN_1_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_1_G₃(x2, x3, x4)
LE_2_IN_GG₂(x1, x2) = LE_2_IN_GG₂(x1, x2)
ORDERED_1_IN_G₁(x1) = ORDERED_1_IN_G₁(x1)
IF_LE_2_IN_1_GG₃(x1, x2, x3) = IF_LE_2_IN_1_GG₁(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:

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_in_gg₂(X, Y))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> IF_ORDERED_1_IN_2_G₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

The TRS R consists of the following rules:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

The TRS R consists of the following rules:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g₁(x1) = ordered_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
ordered_1_out_g₁(x1) = ordered_1_out_g
if_ordered_1_in_1_g₄(x1, x2, x3, x4) = if_ordered_1_in_1_g₃(x2, x3, x4)
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
if_ordered_1_in_2_g₄(x1, x2, x3, x4) = if_ordered_1_in_2_g₁(x4)
LE_2_IN_GG₂(x1, x2) = LE_2_IN_GG₂(x1, 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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LE_2_IN_GG₂}.
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:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(X, []_0)) -> ordered_1_out_g₁(._2₂(X, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_ordered_1_in_1_g₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))

The argument filtering Pi contains the following mapping:
ordered_1_in_g₁(x1) = ordered_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
ordered_1_out_g₁(x1) = ordered_1_out_g
if_ordered_1_in_1_g₄(x1, x2, x3, x4) = if_ordered_1_in_1_g₃(x2, x3, x4)
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
if_ordered_1_in_2_g₄(x1, x2, x3, x4) = if_ordered_1_in_2_g₁(x4)
IF_ORDERED_1_IN_1_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_1_G₃(x2, x3, x4)
ORDERED_1_IN_G₁(x1) = ORDERED_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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_out_gg₂(X, Y)) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg₂(0_0, s_1₁(0_0))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
IF_ORDERED_1_IN_1_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_1_G₃(x2, x3, x4)
ORDERED_1_IN_G₁(x1) = ORDERED_1_IN_G₁(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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

IF_ORDERED_1_IN_1_G₃(Y, Xs, le_2_out_gg) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₃(Y, Xs, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg

The set Q consists of the following terms:

le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {ORDERED_1_IN_G₁, IF_ORDERED_1_IN_1_G₃}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₃(Y, Xs, le_2_in_gg₂(X, Y))

Strictly oriented rules of the TRS R:

le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(0_0)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 0
POL(._2₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(ORDERED_1_IN_G₁(x₁)) = x₁
POL(le_2_out_gg) = 0
POL(le_2_in_gg₂(x₁, x₂)) = 1 + x₁ + x₂
POL(s_1₁(x₁)) = 1 + x₁
POL(if_le_2_in_1_gg₁(x₁)) = 1 + x₁
POL(IF_ORDERED_1_IN_1_G₃(x₁, x₂, x₃)) = 1 + x₁ + 2·x₂ + x₃

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_ORDERED_1_IN_1_G₃(Y, Xs, le_2_out_gg) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

R is empty.
The set Q consists of the following terms:

le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {ORDERED_1_IN_G₁, IF_ORDERED_1_IN_1_G₃}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.