(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:
minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(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:
minus(minus(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:
minus(f(x, y)) → f(minus(y), minus(x))
minus(h(x)) → h(minus(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:
minus(f(x, y)) → f(minus(y), minus(x))
minus(h(x)) → h(minus(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:
f0(0, 0) → 0
h0(0) → 0
minus0(0) → 1
minus1(0) → 2
minus1(0) → 3
f1(2, 3) → 1
minus1(0) → 4
h1(4) → 1
f1(2, 3) → 2
f1(2, 3) → 3
f1(2, 3) → 4
h1(4) → 2
h1(4) → 3
h1(4) → 4
(6) BOUNDS(1, n^1)