(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:
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x
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:
half(double(x)) → x
(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:
double(0) → 0
double(s(x)) → s(s(double(x)))
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(s(x), y, z) → z
half(0) → 0
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
if(0, y, z) → y
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:
double(0) → 0
double(s(x)) → s(s(double(x)))
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(s(x), y, z) → z
half(0) → 0
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
if(0, y, z) → y
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, 4]
transitions:
00() → 0
s0(0) → 0
double0(0) → 1
half0(0) → 2
if0(0, 0, 0) → 3
-0(0, 0) → 4
01() → 1
double1(0) → 6
s1(6) → 5
s1(5) → 1
01() → 2
half1(0) → 7
s1(7) → 2
-1(0, 0) → 4
01() → 6
s1(5) → 6
01() → 7
s1(7) → 7
0 → 3
0 → 4
(6) BOUNDS(1, n^1)