Termination w.r.t. Q proof of AG01

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 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 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPOrderProof (⇔, 13 ms)

        ↳27 QDP

        ↳28 PisEmptyProof (⇔, 0 ms)

        ↳29 YES

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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:

LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) → LE(s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) → LE(y, x)
IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
IF_MOD(true, s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(x1))

(11) Obligation:

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

LE(s(x), s(y)) → LE(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:

LE(s(x), s(y)) → LE(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:

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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:

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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].

minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(x1))

(18) Obligation:

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

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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:

IF_MINUS(false, s(x), y) → MINUS(x, y)
The graph contains the following edges 2 > 1, 3 >= 2
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
The graph contains the following edges 1 >= 2, 2 >= 3

(20) YES

(21) Obligation:

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

IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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:

IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(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].

mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, s(x0), s(x1))
if_mod(false, s(x0), s(x1))

(25) Obligation:

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

IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)

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

(26) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

IF_MOD(true, s(x), s(y)) → MOD(minus(x, y), s(y))
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

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

POL( MOD(x₁, x₂) ) = 2x₁ + 1

POL( minus(x₁, x₂) ) = 2x₁

POL( 0 ) = 0

POL( s(x₁) ) = 2x₁ + 2

POL( if_minus(x₁, ..., x₃) ) = max{0, x₁ + 2x₂ - 2}

POL( le(x₁, x₂) ) = 2

POL( IF_MOD(x₁, ..., x₃) ) = max{0, x₁ + 2x₂ - 2}

POL( true ) = 0

POL( false ) = 0

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

minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(false, s(x), y) → s(minus(x, y))
if_minus(true, s(x), y) → 0

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(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) QDPSizeChangeProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QReductionProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPSizeChangeProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) UsableRulesProof (EQUIVALENT transformation)

(23) Obligation:

(24) QReductionProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPOrderProof (EQUIVALENT transformation)

(27) Obligation:

(28) PisEmptyProof (EQUIVALENT transformation)

(29) YES