Termination w.r.t. Q proof of Secret_06_TRS

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔, 0 ms)

↳2 QDP

↳3 TransformationProof (⇔, 0 ms)

↳4 QDP

↳5 TransformationProof (⇔, 0 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 QDP

↳9 TransformationProof (⇔, 0 ms)

↳10 QDP

↳11 DependencyGraphProof (⇔, 0 ms)

↳12 QDP

↳13 QDPOrderProof (⇔, 68 ms)

↳14 QDP

↳15 TransformationProof (⇔, 0 ms)

↳16 QDP

↳17 TransformationProof (⇔, 0 ms)

↳18 QDP

↳19 DependencyGraphProof (⇔, 0 ms)

↳20 QDP

↳21 QDPOrderProof (⇔, 85 ms)

↳22 QDP

↳23 QDPOrderProof (⇔, 0 ms)

↳24 QDP

↳25 DependencyGraphProof (⇔, 0 ms)

↳26 TRUE

(0) Obligation:

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

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

B(x, y) → A(c(y), a(0, x))
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(3) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule B(x, y) → A(c(y), a(0, x)) we obtained the following new rules [LPAR04]:

B(y_3, 0) → A(c(0), a(0, y_3)) → B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(c(y_0), a(0, 0)) → B(0, y_0) → A(c(y_0), a(0, 0))

(4) Obligation:

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

B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(c(y_0), a(0, 0))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(5) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule B(0, y_0) → A(c(y_0), a(0, 0)) at position [1] we obtained the following new rules [LPAR04]:

B(0, y0) → A(c(y0), 0) → B(0, y0) → A(c(y0), 0)

(6) Obligation:

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

B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y0) → A(c(y0), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

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

A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(9) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule B(x, y) → A(0, x) we obtained the following new rules [LPAR04]:

B(y_3, 0) → A(0, y_3) → B(y_3, 0) → A(0, y_3)
B(0, y_0) → A(0, 0) → B(0, y_0) → A(0, 0)

(10) Obligation:

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

A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
B(0, y_0) → A(0, 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

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

B(y_3, 0) → A(c(0), a(0, y_3))
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( B(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( c(x₁) ) = max{0, x₁ - 1}

POL( a(x₁, x₂) ) = x₁ + x₂

POL( b(x₁, x₂) ) = x₁ + x₂

POL( 0 ) = 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)
b(x, y) → c(a(c(y), a(0, x)))

(14) Obligation:

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

B(y_3, 0) → A(c(0), a(0, y_3))
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(15) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0) we obtained the following new rules [LPAR04]:

A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0) → A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0) → A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)

(16) Obligation:

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

B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → B(0, y)
A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(17) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule A(y, c(b(a(0, x), 0))) → B(0, y) we obtained the following new rules [LPAR04]:

A(c(0), c(b(a(0, x1), 0))) → B(0, c(0)) → A(c(0), c(b(a(0, x1), 0))) → B(0, c(0))
A(0, c(b(a(0, x1), 0))) → B(0, 0) → A(0, c(b(a(0, x1), 0))) → B(0, 0)

(18) Obligation:

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

B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)
A(c(0), c(b(a(0, x1), 0))) → B(0, c(0))
A(0, c(b(a(0, x1), 0))) → B(0, 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(20) Obligation:

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

A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)
A(0, c(b(a(0, x1), 0))) → B(0, 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

A(0, c(b(a(0, x1), 0))) → B(0, 0)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A(x₁, x₂)) = 0A + 0A · x₁ + 0A · x₂

POL(c(x₁)) = 0A + 0A · x₁

POL(0) = 0A

POL(b(x₁, x₂)) = 1A + 0A · x₁ + 0A · x₂

POL(a(x₁, x₂)) = 0A + 0A · x₁ + 0A · x₂

POL(B(x₁, x₂)) = 0A + 0A · x₁ + 0A · x₂

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

(22) Obligation:

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

A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(A(x₁, x₂)) = 0 +
[ 1, 0 ]

· x₁ +
[ 0, 1 ]

· x₂

POL(c(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(0) =
/ 1 \

\ 0 /

POL(b(x₁, x₂)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 0 1 /

· x₁ +
/ 1 1 \

\ 1 1 /

· x₂

POL(a(x₁, x₂)) =
/ 1 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁ +
/ 0 1 \

\ 0 1 /

· x₂

POL(B(x₁, x₂)) = 1 +
[ 0, 1 ]

· x₁ +
[ 1, 0 ]

· x₂

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

(24) Obligation:

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

A(c(0), c(b(a(0, x1), 0))) → B(a(c(b(0, c(0))), x1), 0)
A(0, c(b(a(0, x1), 0))) → B(a(c(b(0, 0)), x1), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) TransformationProof (EQUIVALENT transformation)

(4) Obligation:

(5) TransformationProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) TransformationProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) TransformationProof (EQUIVALENT transformation)

(16) Obligation:

(17) TransformationProof (EQUIVALENT transformation)

(18) Obligation:

(19) DependencyGraphProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) TRUE