(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(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(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, h_1|0, activate_1|0, n__f_1|1, n__h_1|1, g_1|1]
1→3[g_1|1]
1→5[h_1|1, n__h_1|2]
1→6[f_1|1, n__f_1|2]
1→7[g_1|2]
2→2[g_1|0, n__h_1|0, n__f_1|0]
3→4[n__h_1|1]
4→2[n__f_1|1]
5→2[activate_1|1, n__h_1|1, n__f_1|1, g_1|1]
5→5[h_1|1, n__h_1|2]
5→6[f_1|1, n__f_1|2]
5→7[g_1|2]
6→2[activate_1|1, n__h_1|1, n__f_1|1, g_1|1]
6→5[h_1|1, n__h_1|2]
6→6[f_1|1, n__f_1|2]
6→7[g_1|2]
7→8[n__h_1|2]
8→6[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(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0
Tuples:

F(z0) → c
F(z0) → c1
H(z0) → c2
ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c5
S tuples:

F(z0) → c
F(z0) → c1
H(z0) → c2
ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

F, H, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 4 trailing nodes:

F(z0) → c
ACTIVATE(z0) → c5
H(z0) → c2
F(z0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0
Tuples:

ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
S tuples:

ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

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

Removed 2 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0
Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

(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__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = [2]x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(n__f(x1)) = [1] + x1   
POL(n__h(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

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

The set S is empty

(14) BOUNDS(1, 1)