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:

not1(x) -> if3(x, false, true)
and2(x, y) -> if3(x, y, false)
or2(x, y) -> if3(x, true, y)
implies2(x, y) -> if3(x, y, true)
=2(x, x) -> true
=2(x, y) -> if3(x, y, not1(y))
if3(true, x, y) -> x
if3(false, x, y) -> y
if3(x, x, if3(x, false, true)) -> true
=2(x, y) -> if3(x, y, if3(y, false, true))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not1(x) -> if3(x, false, true)
and2(x, y) -> if3(x, y, false)
or2(x, y) -> if3(x, true, y)
implies2(x, y) -> if3(x, y, true)
=2(x, x) -> true
=2(x, y) -> if3(x, y, not1(y))
if3(true, x, y) -> x
if3(false, x, y) -> y
if3(x, x, if3(x, false, true)) -> true
=2(x, y) -> if3(x, y, if3(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:

AND2(x, y) -> IF3(x, y, false)
NOT1(x) -> IF3(x, false, true)
OR2(x, y) -> IF3(x, true, y)
=12(x, y) -> NOT1(y)
=12(x, y) -> IF3(x, y, if3(y, false, true))
=12(x, y) -> IF3(x, y, not1(y))
IMPLIES2(x, y) -> IF3(x, y, true)
=12(x, y) -> IF3(y, false, true)

The TRS R consists of the following rules:

not1(x) -> if3(x, false, true)
and2(x, y) -> if3(x, y, false)
or2(x, y) -> if3(x, true, y)
implies2(x, y) -> if3(x, y, true)
=2(x, x) -> true
=2(x, y) -> if3(x, y, not1(y))
if3(true, x, y) -> x
if3(false, x, y) -> y
if3(x, x, if3(x, false, true)) -> true
=2(x, y) -> if3(x, y, if3(y, false, true))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

AND2(x, y) -> IF3(x, y, false)
NOT1(x) -> IF3(x, false, true)
OR2(x, y) -> IF3(x, true, y)
=12(x, y) -> NOT1(y)
=12(x, y) -> IF3(x, y, if3(y, false, true))
=12(x, y) -> IF3(x, y, not1(y))
IMPLIES2(x, y) -> IF3(x, y, true)
=12(x, y) -> IF3(y, false, true)

The TRS R consists of the following rules:

not1(x) -> if3(x, false, true)
and2(x, y) -> if3(x, y, false)
or2(x, y) -> if3(x, true, y)
implies2(x, y) -> if3(x, y, true)
=2(x, x) -> true
=2(x, y) -> if3(x, y, not1(y))
if3(true, x, y) -> x
if3(false, x, y) -> y
if3(x, x, if3(x, false, true)) -> true
=2(x, y) -> if3(x, y, if3(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.