(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = x1
POL(adx(x1)) = 2·x1
POL(cons(x1, x2)) = x1 + 2·x2
POL(head(x1)) = 1 + x1
POL(incr(x1)) = x1
POL(n__adx(x1)) = 2·x1
POL(n__incr(x1)) = x1
POL(n__zeros) = 0
POL(nats) = 1
POL(nil) = 0
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:
nats → adx(zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(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), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = x1
POL(adx(x1)) = 2·x1
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(incr(x1)) = x1
POL(n__adx(x1)) = 2·x1
POL(n__incr(x1)) = x1
POL(n__zeros) = 0
POL(nil) = 2
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:
adx(nil) → nil
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
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)) → ACTIVATE(L)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__adx(X)) → ADX(X)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
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 1 SCC with 1 less node.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)
The TRS R consists of the following rules:
incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) 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' = [x0 / 0] on the rule
ADX(cons(0, n__zeros))[ ]n[ ] → ADX(cons(0, n__zeros))[ ]n[x0 / 0]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu
ADX(cons(x0, n__zeros))[ ]n[ ] → ADX(cons(0, n__zeros))[ ]n[ ]
by Narrowing at position: [0]
ADX(cons(x0, n__zeros))[ ]n[ ] → ADX(zeros)[ ]n[ ]
by Narrowing at position: [0]
intermediate steps: Instantiation
ADX(cons(x0, x1))[ ]n[ ] → ADX(activate(x1))[ ]n[ ]
by Narrowing at position: []
intermediate steps: Instantiation
ADX(cons(X, L))[ ]n[ ] → INCR(cons(X, n__adx(activate(L))))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Instantiation - Instantiation - Instantiation
INCR(cons(x0, n__adx(y0)))[ ]n[ ] → ADX(y0)[ ]n[ ]
by Narrowing at position: []
intermediate steps: Instantiation - Instantiation
INCR(cons(X, L))[ ]n[ ] → ACTIVATE(L)[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Instantiation
ACTIVATE(n__adx(X))[ ]n[ ] → ADX(X)[ ]n[ ]
by OriginalRule from TRS P
activate(n__zeros)[ ]n[ ] → zeros[ ]n[ ]
by OriginalRule from TRS R
zeros[ ]n[ ] → cons(0, n__zeros)[ ]n[ ]
by OriginalRule from TRS R
(10) NO