Termination w.r.t. Q proof of AProVE_07

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 8 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 UsableRulesReductionPairsProof (⇔, 0 ms)

        ↳13 QDP

        ↳14 PisEmptyProof (⇔, 0 ms)

        ↳15 YES

    ↳16 QDP

        ↳17 UsableRulesProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QReductionProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 QDPOrderProof (⇔, 61 ms)

        ↳22 QDP

        ↳23 DependencyGraphProof (⇔, 0 ms)

        ↳24 TRUE

(0) Obligation:

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

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) → SUM(x, y)
WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
WEIGHT(x) → EMPTY(x)
WEIGHT(x) → EMPTY(tail(x))
WEIGHT(x) → TAIL(x)
IF(false, b, x) → IF2(b, x)
IF2(true, x) → HEAD(x)
IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
IF2(false, x) → SUM(x, cons(0, tail(tail(x))))
IF2(false, x) → TAIL(tail(x))
IF2(false, x) → TAIL(x)

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

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

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

(11) Obligation:

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

SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

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

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0
POL(SUM(x₁, x₂)) = 2·x₁ + x₂
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(s(x₁)) = 2 + x₁

(13) Obligation:

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) YES

(16) Obligation:

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

IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
IF(false, b, x) → IF2(b, x)

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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

(17) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
IF(false, b, x) → IF2(b, x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
sum(cons(0, x), y) → sum(x, y)
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(nil, y) → y

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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

(19) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

(20) Obligation:

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

IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
IF(false, b, x) → IF2(b, x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
sum(cons(0, x), y) → sum(x, y)
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(nil, y) → y

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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.

WEIGHT(x) → IF(empty(x), empty(tail(x)), x)

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

POL(0) = 0
POL(IF(x₁, x₂, x₃)) = [1/4]x₁ + [1/2]x₂ + [1/4]x₃
POL(IF2(x₁, x₂)) = [1] + [1/2]x₁ + [1/4]x₂
POL(WEIGHT(x₁)) = [2] + x₁
POL(cons(x₁, x₂)) = [1] + [2]x₁ + [4]x₂
POL(empty(x₁)) = [2] + [2]x₁
POL(false) = [4]
POL(nil) = 0
POL(s(x₁)) = 0
POL(sum(x₁, x₂)) = x₂
POL(tail(x₁)) = [1/4]x₁
POL(true) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

tail(nil) → nil
tail(cons(n, x)) → x
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
empty(nil) → true
empty(cons(n, x)) → false

(22) Obligation:

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

IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
IF(false, b, x) → IF2(b, x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
sum(cons(0, x), y) → sum(x, y)
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(nil, y) → y

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) QReductionProof (EQUIVALENT transformation)

(11) Obligation:

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(13) Obligation:

(14) PisEmptyProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) QReductionProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) TRUE