(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:
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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:
g(s(f(x))) → g(f(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:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(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:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(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]
transitions:
c0(0, 0) → 0
s0(0) → 0
g0(0) → 1
f0(0) → 2
s1(0) → 4
c1(4, 0) → 3
g1(3) → 1
s1(0) → 6
c1(0, 6) → 5
f1(5) → 2
s1(4) → 4
s1(6) → 6
(6) BOUNDS(1, n^1)