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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) 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))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

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

(7) 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(x4)
DEL(x1)  =  x1
.(x1, x2)  =  .(x2)

Recursive Path Order [RPO].
Precedence:

[F₁, .₁]

The following usable rules [FROCOS05] were oriented: none

(8) 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))

The set Q consists of the following terms:

del(.(x0, .(x1, x2)))
f(true, x0, x1, x2)
f(false, x0, x1, x2)
=(nil, nil)
=(.(x0, x1), nil)
=(nil, .(x0, x1))
=(.(x0, x1), .(u, v))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) TRUE