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

(0) Obligation:

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

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

IMPLIES(false, y) → NOT(false)
IMPLIES(x, false) → NOT(x)
IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))
IMPLIES(not(x), not(y)) → AND(x, y)

The TRS R consists of the following rules:

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))

The TRS R consists of the following rules:

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
IMPLIES(x0, x1, x2)  =  IMPLIES(x0, x2)

Tags:
IMPLIES has argument tags [0,1,0] and root tag 0

Comparison: DMS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
IMPLIES(x1, x2)  =  IMPLIES(x1)
not(x1)  =  not(x1)
and(x1, x2)  =  and(x1)
false  =  false

Recursive path order with status [RPO].
Quasi-Precedence:

[IMPLIES₁, not₁, and₁, false]

Status:

IMPLIES₁: multiset
not₁: multiset
and₁: multiset
false: multiset

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE