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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  F(x3, x4)
true  =  true
DEL(x1)  =  x1
.(x1, x2)  =  .(x1, x2)
=(x1, x2)  =  =(x1, x2)
false  =  false
nil  =  nil
u  =  u
v  =  v
and(x1, x2)  =  and

Lexicographic path order with status [LPO].
Quasi-Precedence:

[true, =₂, v] > [F₂, .₂] > [u, and]
false > [F₂, .₂] > [u, and]

Status:

F₂: [2,1]
true: []
.₂: [2,1]
=₂: [2,1]
false: []
nil: []
u: []
v: []
and: []

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

F(true, x, y, z) → DEL(.(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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(8) TRUE