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:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

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:

DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)
DEL(.(x, .(y, z))) → =¹(x, y)
F(true, x, y, z) → DEL(.(y, z))
F(false, x, y, z) → DEL(.(y, z))
=¹(.(x, y), .(u, v)) → =¹(x, u)
=¹(.(x, y), .(u, v)) → =¹(y, v)

The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

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:

F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)
F(false, x, y, z) → DEL(.(y, z))

The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

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.

F(true, x, y, z) → DEL(.(y, z))
DEL(.(x, .(y, z))) → F(=(x, y), x, y, z)
F(false, x, y, z) → DEL(.(y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

nil > false > [F₄, .₂, v] > [=₂, u, and₂] > [true, DEL₁]

Status:

F₄: [4,3,1,2]
true: multiset
DEL₁: [1]
.₂: [2,1]
=₂: [2,1]
false: multiset
nil: multiset
u: multiset
v: multiset
and₂: multiset

The following usable rules [FROCOS05] were oriented:

=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

(6) Obligation:

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

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

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