YES Termination w.r.t. Q proof of Strategy_removed_AG01_4.29.ari

(0) Obligation:

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(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:

EVEN(s(s(x))) → EVEN(x)
HALF(s(s(x))) → HALF(x)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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:

PLUS(s(x), y) → PLUS(x, y)

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

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(13) YES

(14) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(15) 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.

(16) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(17) 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].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • HALF(s(s(x))) → HALF(x)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(22) 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.

(23) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(24) 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].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(25) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • EVEN(s(s(x))) → EVEN(x)
    The graph contains the following edges 1 > 1

(27) YES

(28) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(29) 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.

(30) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(31) 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].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(32) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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

(33) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF_TIMES(false, s(x), y) → TIMES(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  TIMES(x1)
s(x1)  =  s(x1)
IF_TIMES(x1, x2, x3)  =  IF_TIMES(x2)
even(x1)  =  even
true  =  true
half(x1)  =  x1
false  =  false
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:
[s1, even] > [TIMES1, IFTIMES1]
[s1, even] > true
[false, 0] > [TIMES1, IFTIMES1]
[false, 0] > true

Status:
TIMES1: [1]
s1: [1]
IFTIMES1: [1]
even: []
true: multiset
false: multiset
0: multiset


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

half(s(s(x))) → s(half(x))
half(0) → 0

(34) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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

(35) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( IF_TIMES(x1, ..., x3) ) = max{0, x1 + x2 - 1}

POL( even(x1) ) = 0

POL( s(x1) ) = 2x1 + 2

POL( 0 ) = 0

POL( false ) = 0

POL( TIMES(x1, x2) ) = max{0, x1 - 1}

POL( half(x1) ) = max{0, x1 - 2}

POL( true ) = 0


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

even(s(0)) → false
even(s(s(x))) → even(x)
half(s(s(x))) → s(half(x))
half(0) → 0
even(0) → true

(36) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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

(37) DependencyGraphProof (EQUIVALENT transformation)

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

(38) TRUE