(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
(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:
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(x, y), 0)
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(AP(x1, x2)) = x1
POL(ap(x1, x2)) = 1 + x1 + x2
POL(f) = 0
POL(g) = 0
POL(s) = 0
The following usable rules [FROCOS05] were oriented:
none
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(9) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
AP(
ap(
ap(
g,
x),
y),
ap(
s,
z)) →
AP(
ap(
ap(
g,
x),
y),
ap(
ap(
x,
y),
0)) at position [1] we obtained the following new rules [LPAR04]:
AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(12) Complex Obligation (AND)
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(14) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
Q is empty.
We have to consider all (P,Q,R)-chains.
(16) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(AP(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | -I | -I | -I | | |
\ | -I | -I | -I | / |
| · | x1 | + | / | 0A | -I | 0A | \ |
| | -I | -I | -I | | |
\ | -I | -I | 0A | / |
| · | x2 |
POL(ap(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | 0A | 0A | 0A | | |
\ | 0A | -I | -I | / |
| · | x1 | + | / | 0A | -I | -I | \ |
| | 0A | 0A | 0A | | |
\ | 0A | -I | 0A | / |
| · | x2 |
The following usable rules [FROCOS05] were oriented:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
(17) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(18) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(19) TRUE
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(21) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
The TRS R consists of the following rules:
ap(f, x) → x
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(23) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
AP(
ap(
ap(
g,
f),
x0),
ap(
s,
y2)) evaluates to t =
AP(
ap(
ap(
g,
f),
x0),
ap(
x0,
0))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [y2 / 0, x0 / s]
- Matcher: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from AP(ap(ap(g, f), s), ap(s, 0)) to AP(ap(ap(g, f), s), ap(s, 0)).
(24) FALSE