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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 AND
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE

(0) Obligation:

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

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(forall, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(forsome, p), app(app(cons, x), xs)) → APP(p, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
app(x1, x2)  =  app(x2)
forall  =  forall
cons  =  cons
forsome  =  forsome
and  =  and
true  =  true
false  =  false
or  =  or
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
forall > [app1, cons] > [true, false] > and
forsome > [app1, cons] > [true, false] > and
or > [true, false] > and
nil > and


The following usable rules [FROCOS05] were oriented:

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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))

(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(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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Complex Obligation (AND)

(11) Obligation:

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

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, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(forsome, p), app(app(cons, x), xs)) → APP(app(forsome, p), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x2)
forsome  =  forsome
cons  =  cons
and  =  and
true  =  true
false  =  false
or  =  or
forall  =  forall
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
APP1 > forsome > [true, false, forall] > [app1, cons]
and > [true, false, forall] > [app1, cons]
or > [true, false, forall] > [app1, cons]
nil > [true, false, forall] > [app1, cons]


The following usable rules [FROCOS05] were oriented:

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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))

(13) Obligation:

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

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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.

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) 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)

The TRS R consists of the following rules:

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(forall, p), app(app(cons, x), xs)) → APP(app(forall, p), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x2)
forall  =  forall
cons  =  cons
and  =  and
true  =  true
false  =  false
or  =  or
nil  =  nil
forsome  =  forsome

Lexicographic Path Order [LPO].
Precedence:
[cons, and, true, false, nil] > [app1, forsome] > forall
[cons, and, true, false, nil] > [app1, forsome] > or


The following usable rules [FROCOS05] were oriented:

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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))

(18) Obligation:

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

app(app(and, true), true) → true
app(app(and, true), false) → false
app(app(and, false), true) → false
app(app(and, false), false) → false
app(app(or, true), true) → true
app(app(or, true), false) → true
app(app(or, false), 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, true), false)
app(app(and, false), true)
app(app(and, false), false)
app(app(or, true), true)
app(app(or, true), false)
app(app(or, false), 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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE