(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(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Rewrite Strategy: FULL
(1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)
Converted rc-obligation to irc-obligation.
As the TRS does not nest defined symbols, we have rc = irc.
(2) 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:
f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Rewrite Strategy: INNERMOST
(3) 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:
nil0() → 0
.0(0, 0) → 0
f0(0) → 1
g0(0) → 2
nil1() → 1
nil1() → 3
f1(0) → 4
.1(3, 4) → 1
.1(0, 0) → 6
.1(0, 6) → 5
f1(5) → 1
nil1() → 2
g1(0) → 7
nil1() → 8
.1(7, 8) → 2
.1(0, 0) → 10
.1(10, 0) → 9
g1(9) → 2
nil1() → 4
.1(3, 4) → 4
f1(6) → 4
f1(5) → 4
.1(0, 6) → 6
nil1() → 7
.1(7, 8) → 7
g1(10) → 7
g1(9) → 7
.1(10, 0) → 10
(4) BOUNDS(1, n^1)