Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔, 83 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 NonLoopProof (⇔, 0 ms)
        ↳11 NO
    ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES

(0) Obligation:

Q restricted 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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

Q restricted 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)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(incr(x1)) = x1   
POL(nats) = 2   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

nats → adx(zeros)


(4) Obligation:

Q restricted 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)

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

INCR(cons(X, L)) → INCR(L)
ADX(cons(X, L)) → INCR(cons(X, adx(L)))
ADX(cons(X, L)) → ADX(L)
ZEROS → ZEROS

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)

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 3 SCCs with 1 less node.

(8) Complex Obligation (AND)

(9) Obligation:

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

ZEROS → ZEROS

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)

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

(10) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
ZEROS[ ]n[ ] → ZEROS[ ]n[ ]
This rule is correct for the QDP as the following derivation shows:

ZEROS[ ]n[ ] → ZEROS[ ]n[ ]
    by OriginalRule from TRS P

(11) NO

(12) Obligation:

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

INCR(cons(X, L)) → INCR(L)

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)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • INCR(cons(X, L)) → INCR(L)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

ADX(cons(X, L)) → ADX(L)

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)

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

(16) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ADX(cons(X, L)) → ADX(L)
    The graph contains the following edges 1 > 1

(17) YES