(0) Obligation:
The Runtime Complexity (full) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)
Rewrite Strategy: FULL
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(x, ++(y, z)) → ++(++(x, y), z)
(2) Obligation:
The Runtime Complexity (full) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
++(.(x, y), z) → .(x, ++(y, z))
rev(nil) → nil
++(nil, y) → y
++(x, nil) → x
make(x) → .(x, nil)
Rewrite Strategy: FULL
(3) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)
Converted rc-obligation to irc-obligation.
As the TRS does not nest defined symbols, we have rc = irc.
(4) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
++(.(x, y), z) → .(x, ++(y, z))
rev(nil) → nil
++(nil, y) → y
++(x, nil) → x
make(x) → .(x, nil)
Rewrite Strategy: INNERMOST
(5) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
.0(0, 0) → 0
nil0() → 0
++0(0, 0) → 1
rev0(0) → 2
make0(0) → 3
++1(0, 0) → 4
.1(0, 4) → 1
nil1() → 2
nil1() → 5
.1(0, 5) → 3
.1(0, 4) → 4
0 → 1
0 → 4
(6) BOUNDS(1, n^1)