(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:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: INNERMOST

(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)

The following defined symbols can occur below the 0th argument of top: proper, active
The following defined symbols can occur below the 0th argument of proper: proper, active
The following defined symbols can occur below the 0th argument of active: proper, active

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))

(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:

g(ok(X)) → ok(g(X))
top(ok(X)) → top(active(X))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
top(mark(X)) → top(proper(X))
proper(a) → ok(a)

Rewrite Strategy: INNERMOST

(3) 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: 5
Accept states: [6]
Transitions:
5→6[g_1|0, top_1|0, f_1|0, proper_1|0]
5→7[ok_1|1]
5→8[top_1|1]
5→9[top_1|1]
5→10[mark_1|1]
5→11[ok_1|1]
5→12[ok_1|1]
5→13[top_1|2]
6→6[ok_1|0, active_1|0, mark_1|0, a|0]
7→6[g_1|1]
7→7[ok_1|1]
8→6[active_1|1]
9→6[proper_1|1]
9→12[ok_1|1]
10→6[f_1|1]
10→10[mark_1|1]
10→11[ok_1|1]
11→6[f_1|1]
11→10[mark_1|1]
11→11[ok_1|1]
12→6[a|1]
13→12[active_1|2]

(4) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(ok(z0)) → ok(g(z0))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
f(mark(z0)) → mark(f(z0))
f(ok(z0)) → ok(f(z0))
proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
PROPER(a) → c5
S tuples:

G(ok(z0)) → c(G(z0))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
PROPER(a) → c5
K tuples:none
Defined Rule Symbols:

g, top, f, proper

Defined Pair Symbols:

G, TOP, F, PROPER

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 2 trailing nodes:

PROPER(a) → c5
TOP(ok(z0)) → c1(TOP(active(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(ok(z0)) → ok(g(z0))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
f(mark(z0)) → mark(f(z0))
f(ok(z0)) → ok(f(z0))
proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
S tuples:

G(ok(z0)) → c(G(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
K tuples:none
Defined Rule Symbols:

g, top, f, proper

Defined Pair Symbols:

G, TOP, F

Compound Symbols:

c, c2, c3, c4

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(ok(z0)) → ok(g(z0))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
f(mark(z0)) → mark(f(z0))
f(ok(z0)) → ok(f(z0))
proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

g, top, f, proper

Defined Pair Symbols:

G, F, TOP

Compound Symbols:

c, c3, c4, c2

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(ok(z0)) → ok(g(z0))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
f(mark(z0)) → mark(f(z0))
f(ok(z0)) → ok(f(z0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

G, F, TOP

Compound Symbols:

c, c3, c4, c2

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)) transformation)

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

F(mark(z0)) → c3(F(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = x1   
POL(G(x1)) = 0   
POL(TOP(x1)) = 0   
POL(a) = [1]   
POL(c(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = [1] + x1 + x12 + x13   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

G(ok(z0)) → c(G(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:

F(mark(z0)) → c3(F(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

G, F, TOP

Compound Symbols:

c, c3, c4, c2

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

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

TOP(mark(z0)) → c2(TOP(proper(z0)))
We considered the (Usable) Rules:

proper(a) → ok(a)
And the Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = 0   
POL(G(x1)) = 0   
POL(TOP(x1)) = x1   
POL(a) = 0   
POL(c(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = 0   
POL(proper(x1)) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

G(ok(z0)) → c(G(z0))
F(ok(z0)) → c4(F(z0))
K tuples:

F(mark(z0)) → c3(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

G, F, TOP

Compound Symbols:

c, c3, c4, c2

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

G(ok(z0)) → c(G(z0))
F(ok(z0)) → c4(F(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = x1 + [2]x12   
POL(G(x1)) = x1 + x12   
POL(TOP(x1)) = 0   
POL(a) = [2]   
POL(c(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = [1] + [2]x12   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(a) → ok(a)
Tuples:

G(ok(z0)) → c(G(z0))
F(mark(z0)) → c3(F(z0))
F(ok(z0)) → c4(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:none
K tuples:

F(mark(z0)) → c3(F(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)))
G(ok(z0)) → c(G(z0))
F(ok(z0)) → c4(F(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

G, F, TOP

Compound Symbols:

c, c3, c4, c2

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

The set S is empty

(20) BOUNDS(1, 1)