Termination w.r.t. Q proof of /tmp/tmplhrjvK/filliatre2.xml

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 DependencyGraphProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 DependencyGraphProof (⇔)

↳16 TRUE

(0) Obligation:

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

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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:

FOLDB(t, s(n)) → F(foldB(t, n), B)
FOLDB(t, s(n)) → FOLDB(t, n)
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)
F(t, x) → F'(t, g(x))
F(t, x) → G(x)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Obligation:

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

F(t, x) → F'(t, g(x))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
FOLDB(t, s(n)) → F(foldB(t, n), B)
FOLDB(t, s(n)) → FOLDB(t, n)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FOLDB(t, s(n)) → F(foldB(t, n), B)
FOLDB(t, s(n)) → FOLDB(t, n)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A) = 0
POL(B) = 0
POL(C) = 0
POL(F(x₁, x₂)) = x₁
POL(F'(x₁, x₂)) = x₁
POL(F''(x₁)) = x₁
POL(FOLDB(x₁, x₂)) = x₁ + x₂
POL(FOLDC(x₁, x₂)) = x₁
POL(f(x₁, x₂)) = x₁
POL(f'(x₁, x₂)) = x₁
POL(f''(x₁)) = x₁
POL(foldB(x₁, x₂)) = x₁
POL(foldC(x₁, x₂)) = x₁
POL(g(x₁)) = 0
POL(s(x₁)) = 1 + x₁
POL(triple(x₁, x₂, x₃)) = x₂

The following usable rules [FROCOS05] were oriented:

foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)
f'(triple(a, b, c), C) → triple(a, b, s(c))

(6) Obligation:

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

F(t, x) → F'(t, g(x))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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:

F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F(t, x) → F'(t, g(x))
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A) = 0
POL(B) = 0
POL(C) = 0
POL(F(x₁, x₂)) = x₁ + x₂
POL(F'(x₁, x₂)) = x₁
POL(F''(x₁)) = x₁
POL(FOLDC(x₁, x₂)) = x₁ + x₂
POL(f(x₁, x₂)) = 1 + x₁
POL(f'(x₁, x₂)) = 1 + x₁
POL(f''(x₁)) = 1 + x₁
POL(foldB(x₁, x₂)) = x₁ + x₂
POL(foldC(x₁, x₂)) = x₁ + x₂
POL(g(x₁)) = 0
POL(s(x₁)) = 1 + x₁
POL(triple(x₁, x₂, x₃)) = x₂ + x₃

The following usable rules [FROCOS05] were oriented:

foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)
f'(triple(a, b, c), C) → triple(a, b, s(c))

(10) Obligation:

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

F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F(t, x) → F'(t, g(x))
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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 2 less nodes.

(12) Obligation:

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

F(t, x) → F'(t, g(x))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A) = 0
POL(B) = 1
POL(C) = 1
POL(F(x₁, x₂)) = x₂
POL(F'(x₁, x₂)) = x₂
POL(g(x₁)) = x₁
POL(triple(x₁, x₂, x₃)) = 0

The following usable rules [FROCOS05] were oriented:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C

(14) Obligation:

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

F(t, x) → F'(t, g(x))

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) TRUE