YES Termination w.r.t. Q proof of AG01_3.5a.ari

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 (⇔, 25 ms)
        ↳27 QDP
        ↳28 DependencyGraphProof (⇔, 0 ms)
        ↳29 TRUE

(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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(x, s(y)) → PRED(minus(x, y))
MINUS(x, s(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

MINUS(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

MINUS(x, s(y)) → MINUS(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

MINUS(x, s(y)) → MINUS(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:

  • MINUS(x, s(y)) → MINUS(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:

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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

LE(s(x), s(y)) → LE(x, y)

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:

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

(20) YES

(21) Obligation:

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

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))

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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(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))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

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))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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:

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))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
pred(s(x0))
minus(x0, 0)
minus(x0, s(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))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MOD(x1, x2)  =  x1
s(x1)  =  s(x1)
IF_MOD(x1, x2, x3)  =  x2
minus(x1, x2)  =  x1
pred(x1)  =  x1

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
dummyConstant=1

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

minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) TRUE