Termination w.r.t. Q proof of TRS/SK902.32.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

not(x) → if(x, false, true)
and(x, y) → if(x, y, false)
or(x, y) → if(x, true, y)
implies(x, y) → if(x, y, true)
=(x, x) → true
=(x, y) → if(x, y, not(y))
if(true, x, y) → x
if(false, x, y) → y
if(x, x, if(x, false, true)) → true
=(x, y) → if(x, y, if(y, false, true))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

not(x) → if(x, false, true)
and(x, y) → if(x, y, false)
or(x, y) → if(x, true, y)
implies(x, y) → if(x, y, true)
=(x, x) → true
=(x, y) → if(x, y, not(y))
if(true, x, y) → x
if(false, x, y) → y
if(x, x, if(x, false, true)) → true
=(x, y) → if(x, y, if(y, false, true))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AND(x, y) → IF(x, y, false)
NOT(x) → IF(x, false, true)
OR(x, y) → IF(x, true, y)
=¹(x, y) → NOT(y)
=¹(x, y) → IF(x, y, if(y, false, true))
=¹(x, y) → IF(x, y, not(y))
IMPLIES(x, y) → IF(x, y, true)
=¹(x, y) → IF(y, false, true)

The TRS R consists of the following rules:

not(x) → if(x, false, true)
and(x, y) → if(x, y, false)
or(x, y) → if(x, true, y)
implies(x, y) → if(x, y, true)
=(x, x) → true
=(x, y) → if(x, y, not(y))
if(true, x, y) → x
if(false, x, y) → y
if(x, x, if(x, false, true)) → true
=(x, y) → if(x, y, if(y, false, true))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

AND(x, y) → IF(x, y, false)
NOT(x) → IF(x, false, true)
OR(x, y) → IF(x, true, y)
=¹(x, y) → NOT(y)
=¹(x, y) → IF(x, y, if(y, false, true))
=¹(x, y) → IF(x, y, not(y))
IMPLIES(x, y) → IF(x, y, true)
=¹(x, y) → IF(y, false, true)

The TRS R consists of the following rules:

not(x) → if(x, false, true)
and(x, y) → if(x, y, false)
or(x, y) → if(x, true, y)
implies(x, y) → if(x, y, true)
=(x, x) → true
=(x, y) → if(x, y, not(y))
if(true, x, y) → x
if(false, x, y) → y
if(x, x, if(x, false, true)) → true
=(x, y) → if(x, y, if(y, false, true))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

NOT(x) → IF(x, false, true)
AND(x, y) → IF(x, y, false)
OR(x, y) → IF(x, true, y)
=¹(x, y) → NOT(y)
=¹(x, y) → IF(x, y, if(y, false, true))
=¹(x, y) → IF(x, y, not(y))
IMPLIES(x, y) → IF(x, y, true)
=¹(x, y) → IF(y, false, true)

The TRS R consists of the following rules:

not(x) → if(x, false, true)
and(x, y) → if(x, y, false)
or(x, y) → if(x, true, y)
implies(x, y) → if(x, y, true)
=(x, x) → true
=(x, y) → if(x, y, not(y))
if(true, x, y) → x
if(false, x, y) → y
if(x, x, if(x, false, true)) → true
=(x, y) → if(x, y, if(y, false, true))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 8 less nodes.