Termination w.r.t. Q proof of /tmp/tmpnOFEJX/13.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)

The signature Sigma is {cond1, cond2}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

COND2(true, x, y) → GR(x, 0)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → P(x)
COND2(true, x, y) → COND1(gr(x, 0), y, y)
GR(s(x), s(y)) → GR(x, y)
COND1(true, x, y) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

COND2(true, x, y) → GR(x, 0)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → P(x)
COND2(true, x, y) → COND1(gr(x, 0), y, y)
GR(s(x), s(y)) → GR(x, y)
COND1(true, x, y) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

GR(s(x), s(y)) → GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(true, x, y) → COND1(gr(x, 0), y, y)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(true, x, y) → COND1(gr(x, 0), y, y)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(true, x, y) → COND1(gr(x, 0), y, y)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.

For Pair COND1(true, x, y) → COND2(gr(x, y), x, y) the following chains were created:

We consider the chain COND2(true, x, y) → COND1(gr(x, 0), y, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(1)    (COND1(gr(x2, 0), x3, x3)=COND1(true, x4, x5) ⇒ COND1(true, x4, x5)≥COND2(gr(x4, x5), x4, x5))

We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
(2)    (0=x26∧gr(x2, x26)=true ⇒ COND1(true, x3, x3)≥COND2(gr(x3, x3), x3, x3))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x2, x26)=true which results in the following new constraints:
(3)    (true=true∧0=0 ⇒ COND1(true, x3, x3)≥COND2(gr(x3, x3), x3, x3))

(4)    (gr(x29, x30)=true∧0=s(x30)∧(∀x31:gr(x29, x30)=true∧0=x30 ⇒ COND1(true, x31, x31)≥COND2(gr(x31, x31), x31, x31)) ⇒ COND1(true, x3, x3)≥COND2(gr(x3, x3), x3, x3))

We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5)    (COND1(true, x3, x3)≥COND2(gr(x3, x3), x3, x3))

We solved constraint (4) using rules (I), (II).
We consider the chain COND2(false, x, y) → COND1(gr(x, 0), p(x), y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(6)    (COND1(gr(x6, 0), p(x6), x7)=COND1(true, x8, x9) ⇒ COND1(true, x8, x9)≥COND2(gr(x8, x9), x8, x9))

We simplified constraint (6) using rules (I), (II), (III), (VII) which results in the following new constraint:
(7)    (0=x32∧gr(x6, x32)=true∧p(x6)=x8 ⇒ COND1(true, x8, x7)≥COND2(gr(x8, x7), x8, x7))

We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on gr(x6, x32)=true which results in the following new constraints:
(8)    (true=true∧0=0∧p(s(x34))=x8 ⇒ COND1(true, x8, x7)≥COND2(gr(x8, x7), x8, x7))

(9)    (gr(x35, x36)=true∧0=s(x36)∧p(s(x35))=x8∧(∀x37,x38:gr(x35, x36)=true∧0=x36∧p(x35)=x37 ⇒ COND1(true, x37, x38)≥COND2(gr(x37, x38), x37, x38)) ⇒ COND1(true, x8, x7)≥COND2(gr(x8, x7), x8, x7))

We simplified constraint (8) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(10)    (COND1(true, x8, x7)≥COND2(gr(x8, x7), x8, x7))

We solved constraint (9) using rules (I), (II).

For Pair COND2(true, x, y) → COND1(gr(x, 0), y, y) the following chains were created:

We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND1(gr(x, 0), y, y) which results in the following constraint:
(11)    (COND2(gr(x10, x11), x10, x11)=COND2(true, x12, x13) ⇒ COND2(true, x12, x13)≥COND1(gr(x12, 0), x13, x13))

We simplified constraint (11) using rules (I), (II), (III) which results in the following new constraint:
(12)    (gr(x10, x11)=true ⇒ COND2(true, x10, x11)≥COND1(gr(x10, 0), x11, x11))

We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on gr(x10, x11)=true which results in the following new constraints:
(13)    (true=true ⇒ COND2(true, s(x41), 0)≥COND1(gr(s(x41), 0), 0, 0))

(14)    (gr(x42, x43)=true∧(gr(x42, x43)=true ⇒ COND2(true, x42, x43)≥COND1(gr(x42, 0), x43, x43)) ⇒ COND2(true, s(x42), s(x43))≥COND1(gr(s(x42), 0), s(x43), s(x43)))

We simplified constraint (13) using rules (I), (II) which results in the following new constraint:
(15)    (COND2(true, s(x41), 0)≥COND1(gr(s(x41), 0), 0, 0))

We simplified constraint (14) using rule (VI) where we applied the induction hypothesis (gr(x42, x43)=true ⇒ COND2(true, x42, x43)≥COND1(gr(x42, 0), x43, x43)) with σ = [ ] which results in the following new constraint:
(16)    (COND2(true, x42, x43)≥COND1(gr(x42, 0), x43, x43) ⇒ COND2(true, s(x42), s(x43))≥COND1(gr(s(x42), 0), s(x43), s(x43)))

For Pair COND2(false, x, y) → COND1(gr(x, 0), p(x), y) the following chains were created:

We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND1(gr(x, 0), p(x), y) which results in the following constraint:
(17)    (COND2(gr(x18, x19), x18, x19)=COND2(false, x20, x21) ⇒ COND2(false, x20, x21)≥COND1(gr(x20, 0), p(x20), x21))

We simplified constraint (17) using rules (I), (II), (III) which results in the following new constraint:
(18)    (gr(x18, x19)=false ⇒ COND2(false, x18, x19)≥COND1(gr(x18, 0), p(x18), x19))

We simplified constraint (18) using rule (V) (with possible (I) afterwards) using induction on gr(x18, x19)=false which results in the following new constraints:
(19)    (false=false ⇒ COND2(false, 0, x44)≥COND1(gr(0, 0), p(0), x44))

(20)    (gr(x46, x47)=false∧(gr(x46, x47)=false ⇒ COND2(false, x46, x47)≥COND1(gr(x46, 0), p(x46), x47)) ⇒ COND2(false, s(x46), s(x47))≥COND1(gr(s(x46), 0), p(s(x46)), s(x47)))

We simplified constraint (19) using rules (I), (II) which results in the following new constraint:
(21)    (COND2(false, 0, x44)≥COND1(gr(0, 0), p(0), x44))

We simplified constraint (20) using rule (VI) where we applied the induction hypothesis (gr(x46, x47)=false ⇒ COND2(false, x46, x47)≥COND1(gr(x46, 0), p(x46), x47)) with σ = [ ] which results in the following new constraint:
(22)    (COND2(false, x46, x47)≥COND1(gr(x46, 0), p(x46), x47) ⇒ COND2(false, s(x46), s(x47))≥COND1(gr(s(x46), 0), p(s(x46)), s(x47)))

To summarize, we get the following constraints P_≥ for the following pairs.

COND1(true, x, y) → COND2(gr(x, y), x, y)
- (COND1(true, x3, x3)≥COND2(gr(x3, x3), x3, x3))
- (COND1(true, x8, x7)≥COND2(gr(x8, x7), x8, x7))
COND2(true, x, y) → COND1(gr(x, 0), y, y)
- (COND2(true, s(x41), 0)≥COND1(gr(s(x41), 0), 0, 0))
- (COND2(true, x42, x43)≥COND1(gr(x42, 0), x43, x43) ⇒ COND2(true, s(x42), s(x43))≥COND1(gr(s(x42), 0), s(x43), s(x43)))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)
- (COND2(false, 0, x44)≥COND1(gr(0, 0), p(0), x44))
- (COND2(false, x46, x47)≥COND1(gr(x46, 0), p(x46), x47) ⇒ COND2(false, s(x46), s(x47))≥COND1(gr(s(x46), 0), p(s(x46)), s(x47)))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [18]:

POL(0) = 0
POL(COND1(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃
POL(COND2(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃
POL(c) = -1
POL(false) = 0
POL(gr(x₁, x₂)) = 0
POL(p(x₁)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

COND2(true, x, y) → COND1(gr(x, 0), y, y)

The following pairs are in P_bound:

COND2(true, x, y) → COND1(gr(x, 0), y, y)

The following rules are usable:

false → gr(0, x)
p(0) → 0
p(s(x)) → x
true → gr(s(x), 0)
gr(x, y) → gr(s(x), s(y))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The remaining pairs can at least be oriented weakly.

COND1(true, x, y) → COND2(gr(x, y), x, y)

Used ordering: Polynomial interpretation [25,35]:

POL(gr(x₁, x₂)) = (4)x_1
POL(true) = 1/2
POL(COND2(x₁, x₂, x₃)) = 4 + (11/4)x_2
POL(false) = 0
POL(p(x₁)) = (1/4)x_1
POL(s(x₁)) = 4 + (4)x_1
POL(0) = 0
POL(COND1(x₁, x₂, x₃)) = 15/4 + (1/2)x_1 + (3)x_2

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

gr(0, x) → false
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

COND1(true, x, y) → COND2(gr(x, y), x, y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.