(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(s(x)), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), s(y)) → PLUS(y, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
The set Q consists of the following terms:
plus(s(s(x0)), x1)
plus(x0, s(s(x1)))
plus(s(0), x0)
plus(0, x0)
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))
R is empty.
The set Q consists of the following terms:
plus(s(s(x0)), x1)
plus(x0, s(s(x1)))
plus(s(0), x0)
plus(0, x0)
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
plus(s(s(x0)), x1)
plus(x0, s(s(x1)))
plus(s(0), x0)
plus(0, x0)
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
PLUS(s(s(x)), y) → PLUS(x, s(y))
Used ordering: Polynomial interpretation [POLO]:
POL(PLUS(x1, x2)) = 2·x1 + x2
POL(s(x1)) = 1 + x1
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(s(y))) → PLUS(s(x), y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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:
- PLUS(x, s(s(y))) → PLUS(s(x), y)
The graph contains the following edges 2 > 2
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
The TRS R consists of the following rules:
plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))
The set Q consists of the following terms:
plus(s(s(x0)), x1)
plus(x0, s(s(x1)))
plus(s(0), x0)
plus(0, x0)
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(19) 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:
- ACK(s(x), 0) → ACK(x, s(0))
The graph contains the following edges 1 > 1
- ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
The graph contains the following edges 1 > 1
- ACK(s(x), s(y)) → ACK(s(x), y)
The graph contains the following edges 1 >= 1, 2 > 2
(20) TRUE