(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:

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))

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:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))

(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:

f(b(x), b(y)) → b(f(x, 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:

f(b(x), b(y)) → b(f(x, 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]
transitions:
b0(0) → 0
f0(0, 0) → 1
f1(0, 0) → 2
b1(2) → 1
b1(2) → 2

(6) BOUNDS(1, n^1)