(0) 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(n__f(n__a)) → f(n__g(n__f(n__a)))
f(X) → n__f(X)
an__a
g(X) → n__g(X)
activate(n__f(X)) → f(X)
activate(n__a) → a
activate(n__g(X)) → g(activate(X))
activate(X) → X

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[f_1|0, a|0, g_1|0, activate_1|0, n__f_1|1, n__a|1, n__g_1|1, f_1|1, a|1, n__f_1|2, n__a|2]
1→3[f_1|1, n__f_1|2]
1→6[g_1|1, n__g_1|2]
2→2[n__f_1|0, n__a|0, n__g_1|0]
3→4[n__g_1|1]
4→5[n__f_1|1]
5→2[n__a|1]
6→2[activate_1|1, f_1|1, n__f_1|1, a|1, n__a|1, n__g_1|1, n__f_1|2, n__a|2]
6→6[g_1|1, n__g_1|2]
6→3[f_1|1, n__f_1|2]

(2) BOUNDS(1, n^1)

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(n__f(n__a)) → f(n__g(n__f(n__a)))
f(z0) → n__f(z0)
an__a
g(z0) → n__g(z0)
activate(n__f(z0)) → f(z0)
activate(n__a) → a
activate(n__g(z0)) → g(activate(z0))
activate(z0) → z0
Tuples:

F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
F(z0) → c1
Ac2
G(z0) → c3
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c7
S tuples:

F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
F(z0) → c1
Ac2
G(z0) → c3
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c7
K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

F, A, G, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 7 trailing nodes:

ACTIVATE(z0) → c7
F(z0) → c1
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__f(z0)) → c4(F(z0))
F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
G(z0) → c3
Ac2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(n__f(n__a)) → f(n__g(n__f(n__a)))
f(z0) → n__f(z0)
an__a
g(z0) → n__g(z0)
activate(n__f(z0)) → f(z0)
activate(n__a) → a
activate(n__g(z0)) → g(activate(z0))
activate(z0) → z0
Tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))
S tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(n__f(n__a)) → f(n__g(n__f(n__a)))
f(z0) → n__f(z0)
an__a
g(z0) → n__g(z0)
activate(n__f(z0)) → f(z0)
activate(n__a) → a
activate(n__g(z0)) → g(activate(z0))
activate(z0) → z0
Tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
S tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(n__f(n__a)) → f(n__g(n__f(n__a)))
f(z0) → n__f(z0)
an__a
g(z0) → n__g(z0)
activate(n__f(z0)) → f(z0)
activate(n__a) → a
activate(n__g(z0)) → g(activate(z0))
activate(z0) → z0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
S tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = x1   
POL(c6(x1)) = x1   
POL(n__g(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(14) BOUNDS(1, 1)