(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:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
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:
++(++(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))
++(nil, y) → y
++(x, nil) → x
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))
++(nil, y) → y
++(x, nil) → x
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]
transitions:
.0(0, 0) → 0
nil0() → 0
++0(0, 0) → 1
++1(0, 0) → 2
.1(0, 2) → 1
.1(0, 2) → 2
0 → 1
0 → 2
(6) BOUNDS(1, n^1)