YES Termination w.r.t. Q proof of AotoYamada_05_012.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 QDP
↳7 UsableRulesProof (⇔, 0 ms)
↳8 QDP
↳9 QDPSizeChangeProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

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:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

app(app(and, true), true)
app(app(and, x0), false)
app(app(and, false), x0)
app(app(or, true), x0)
app(app(or, x0), true)
app(app(or, false), false)
app(app(forall, x0), nil)
app(app(forall, x0), app(app(cons, x1), x2))
app(app(forsome, x0), nil)
app(app(forsome, x0), app(app(cons, x1), x2))

(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:

APP(app(forall, p), app(app(cons, x), xs)) → APP(app(and, app(p, x)), app(app(forall, p), xs))
APP(app(forall, p), app(app(cons, x), xs)) → APP(and, app(p, x))
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(or, app(p, x)), app(app(forsome, p), xs))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(or, app(p, x))
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

app(app(and, true), true)
app(app(and, x0), false)
app(app(and, false), x0)
app(app(or, true), x0)
app(app(or, x0), true)
app(app(or, false), false)
app(app(forall, x0), nil)
app(app(forall, x0), app(app(cons, x1), x2))
app(app(forsome, x0), nil)
app(app(forsome, x0), app(app(cons, x1), x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, x), false) → false
app(app(and, false), y) → false
app(app(or, true), y) → true
app(app(or, x), true) → true
app(app(or, false), false) → false
app(app(forall, p), nil) → true
app(app(forall, p), app(app(cons, x), xs)) → app(app(and, app(p, x)), app(app(forall, p), xs))
app(app(forsome, p), nil) → false
app(app(forsome, p), app(app(cons, x), xs)) → app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

app(app(and, true), true)
app(app(and, x0), false)
app(app(and, false), x0)
app(app(or, true), x0)
app(app(or, x0), true)
app(app(or, false), false)
app(app(forall, x0), nil)
app(app(forall, x0), app(app(cons, x1), x2))
app(app(forsome, x0), nil)
app(app(forsome, x0), app(app(cons, x1), x2))

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

(7) 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.

(8) Obligation:

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

APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)

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

app(app(and, true), true)
app(app(and, x0), false)
app(app(and, false), x0)
app(app(or, true), x0)
app(app(or, x0), true)
app(app(or, false), false)
app(app(forall, x0), nil)
app(app(forall, x0), app(app(cons, x1), x2))
app(app(forsome, x0), nil)
app(app(forsome, x0), app(app(cons, x1), x2))

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

(9) 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:

  • APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

(10) YES