(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(adx(x1)) = x1
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(hd(x1)) = 2 + 2·x1
POL(incr(x1)) = x1
POL(nats) = 2
POL(s(x1)) = x1
POL(tl(x1)) = 1 + 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)
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
Q is empty.
(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:
ZEROS → ZEROS
INCR(cons(X, Y)) → INCR(Y)
ADX(cons(X, Y)) → INCR(cons(X, adx(Y)))
ADX(cons(X, Y)) → ADX(Y)
The TRS R consists of the following rules:
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INCR(cons(X, Y)) → INCR(Y)
The TRS R consists of the following rules:
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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, Y)) → INCR(Y)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADX(cons(X, Y)) → ADX(Y)
The TRS R consists of the following rules:
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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, Y)) → ADX(Y)
The graph contains the following edges 1 > 1
(12) YES
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ZEROS → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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
(15) NO