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:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) QTRSToCSRProof (EQUIVALENT transformation)

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top, active: active, mark: mark, ok: ok, proper: proper
The replacement map contains the following entries:

incr: {1}
nil: empty set
cons: {1}
s: {1}
adx: {1}
nats: empty set
zeros: empty set
0: empty set
head: {1}
tail: {1}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).

(2) Obligation:

Context-sensitive rewrite system:
The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The replacement map contains the following entries:

incr: {1}
nil: empty set
cons: {1}
s: {1}
adx: {1}
nats: empty set
zeros: empty set
0: empty set
head: {1}
tail: {1}

(3) PoloCSRProof (EQUIVALENT transformation)

The following rules can be removed because they are oriented strictly by a µ-monotonic polynomial ordering:

adx(nil) → nil
head(cons(X, L)) → X
tail(cons(X, L)) → L

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁ + x₂   
POL(head(x₁)) = 2 + x₁   
POL(incr(x₁)) = x₁   
POL(nats) = 0   
POL(nil) = 2   
POL(s(x₁)) = x₁   
POL(tail(x₁)) = 2 + x₁   
POL(zeros) = 0   

(4) Obligation:

Context-sensitive rewrite system:
The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)

The replacement map contains the following entries:

incr: {1}
nil: empty set
cons: {1}
s: {1}
adx: {1}
nats: empty set
zeros: empty set
0: empty set

(5) PoloCSRProof (EQUIVALENT transformation)

The following rules can be removed because they are oriented strictly by a µ-monotonic polynomial ordering:

nats → adx(zeros)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁   
POL(incr(x₁)) = 2·x₁   
POL(nats) = 2   
POL(nil) = 0   
POL(s(x₁)) = 2·x₁   
POL(zeros) = 0   

(6) Obligation:

Context-sensitive rewrite system:
The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The replacement map contains the following entries:

incr: {1}
nil: empty set
cons: {1}
s: {1}
adx: {1}
zeros: empty set
0: empty set

(7) PoloCSRProof (EQUIVALENT transformation)

The following rules can be removed because they are oriented strictly by a µ-monotonic polynomial ordering:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x₁)) = 2 + 2·x₁   
POL(cons(x₁, x₂)) = 1 + x₁   
POL(incr(x₁)) = 1 + 2·x₁   
POL(nil) = 2   
POL(s(x₁)) = 1 + x₁   
POL(zeros) = 2   

(8) Obligation:

Context-sensitive rewrite system:
R is empty.

(9) RisEmptyProof (EQUIVALENT transformation)

The CSR R is empty. Hence, termination is trivially proven.

(10) TRUE