The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

0 CpxTRS
↳1 NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 221 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳6 CdtProblem
        ↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 CdtProblem
        ↳11 CdtUsableRulesProof (⇔, 0 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 75 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 15 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 16 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳20 CdtProblem
        ↳21 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 9 ms)
        ↳22 CdtProblem
        ↳23 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳24 BOUNDS(1, 1)

(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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))

(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(mark(X1), X2) → mark(f(X1, X2))
top(ok(X)) → top(active(X))
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
active(a) → mark(b)
top(mark(X)) → top(proper(X))
h(mark(X)) → mark(h(X))
proper(a) → ok(a)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))

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 4.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3, 4, 5, 6]
transitions:
mark0(0) → 0
ok0(0) → 0
b0() → 0
a0() → 0
f0(0, 0) → 1
top0(0) → 2
proper0(0) → 3
g0(0, 0) → 4
active0(0) → 5
h0(0) → 6
f1(0, 0) → 7
mark1(7) → 1
active1(0) → 8
top1(8) → 2
b1() → 9
ok1(9) → 3
f1(0, 0) → 10
ok1(10) → 1
g1(0, 0) → 11
mark1(11) → 4
b1() → 12
mark1(12) → 5
proper1(0) → 13
top1(13) → 2
h1(0) → 14
mark1(14) → 6
a1() → 15
ok1(15) → 3
h1(0) → 16
ok1(16) → 6
g1(0, 0) → 17
ok1(17) → 4
mark1(7) → 7
mark1(7) → 10
ok1(9) → 13
ok1(10) → 7
ok1(10) → 10
mark1(11) → 11
mark1(11) → 17
mark1(12) → 8
mark1(14) → 14
mark1(14) → 16
ok1(15) → 13
ok1(16) → 14
ok1(16) → 16
ok1(17) → 11
ok1(17) → 17
active2(9) → 18
top2(18) → 2
active2(15) → 18
proper2(12) → 19
top2(19) → 2
b2() → 20
ok2(20) → 19
b2() → 21
mark2(21) → 18
active3(20) → 22
top3(22) → 2
proper3(21) → 23
top3(23) → 2
b3() → 24
ok3(24) → 23
active4(24) → 25
top4(25) → 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:

f(mark(z0), z1) → mark(f(z0, z1))
f(ok(z0), ok(z1)) → ok(f(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(a) → ok(a)
g(mark(z0), z1) → mark(g(z0, z1))
g(ok(z0), ok(z1)) → ok(g(z0, z1))
active(a) → mark(b)
h(mark(z0)) → mark(h(z0))
h(ok(z0)) → ok(h(z0))
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
TOP(ok(z0)) → c2(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c4
PROPER(a) → c5
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
ACTIVE(a) → c8
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
S tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
TOP(ok(z0)) → c2(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c4
PROPER(a) → c5
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
ACTIVE(a) → c8
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
K tuples:none
Defined Rule Symbols:

f, top, proper, g, active, h

Defined Pair Symbols:

F, TOP, PROPER, G, ACTIVE, H

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10

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

Removed 3 trailing nodes:

ACTIVE(a) → c8
PROPER(a) → c5
PROPER(b) → c4

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(mark(z0), z1) → mark(f(z0, z1))
f(ok(z0), ok(z1)) → ok(f(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(a) → ok(a)
g(mark(z0), z1) → mark(g(z0, z1))
g(ok(z0), ok(z1)) → ok(g(z0, z1))
active(a) → mark(b)
h(mark(z0)) → mark(h(z0))
h(ok(z0)) → ok(h(z0))
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
TOP(ok(z0)) → c2(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)), PROPER(z0))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
S tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
TOP(ok(z0)) → c2(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)), PROPER(z0))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
K tuples:none
Defined Rule Symbols:

f, top, proper, g, active, h

Defined Pair Symbols:

F, TOP, G, H

Compound Symbols:

c, c1, c2, c3, c6, c7, c9, c10

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

Removed 2 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(mark(z0), z1) → mark(f(z0, z1))
f(ok(z0), ok(z1)) → ok(f(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(a) → ok(a)
g(mark(z0), z1) → mark(g(z0, z1))
g(ok(z0), ok(z1)) → ok(g(z0, z1))
active(a) → mark(b)
h(mark(z0)) → mark(h(z0))
h(ok(z0)) → ok(h(z0))
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

f, top, proper, g, active, h

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(mark(z0), z1) → mark(f(z0, z1))
f(ok(z0), ok(z1)) → ok(f(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
g(mark(z0), z1) → mark(g(z0, z1))
g(ok(z0), ok(z1)) → ok(g(z0, z1))
h(mark(z0)) → mark(h(z0))
h(ok(z0)) → ok(h(z0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

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

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

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = [2]x2   
POL(G(x1, x2)) = [2]x2   
POL(H(x1)) = 0   
POL(TOP(x1)) = 0   
POL(a) = [2]   
POL(active(x1)) = [2] + [3]x1   
POL(b) = [2]   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [3]   
POL(ok(x1)) = [3] + x1   
POL(proper(x1)) = [2] + [3]x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

F(mark(z0), z1) → c(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
K tuples:

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

(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.

F(mark(z0), z1) → c(F(z0, z1))
H(mark(z0)) → c9(H(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)))
We considered the (Usable) Rules:

proper(b) → ok(b)
active(a) → mark(b)
proper(a) → ok(a)
And the Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x1   
POL(G(x1, x2)) = 0   
POL(H(x1)) = x1   
POL(TOP(x1)) = x1   
POL(a) = [1]   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

G(mark(z0), z1) → c6(G(z0, z1))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
K tuples:

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
F(mark(z0), z1) → c(F(z0, z1))
H(mark(z0)) → c9(H(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

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

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

G(mark(z0), z1) → c6(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = 0   
POL(G(x1, x2)) = x1 + x2   
POL(H(x1)) = 0   
POL(TOP(x1)) = 0   
POL(a) = 0   
POL(active(x1)) = 0   
POL(b) = [1]   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = [1]   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
K tuples:

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
F(mark(z0), z1) → c(F(z0, z1))
H(mark(z0)) → c9(H(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)))
G(mark(z0), z1) → c6(G(z0, z1))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

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

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

H(ok(z0)) → c10(H(z0))
We considered the (Usable) Rules:none
And the Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = 0   
POL(G(x1, x2)) = 0   
POL(H(x1)) = x1   
POL(TOP(x1)) = 0   
POL(a) = [1]   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = x1   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:

TOP(ok(z0)) → c2(TOP(active(z0)))
K tuples:

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
F(mark(z0), z1) → c(F(z0, z1))
H(mark(z0)) → c9(H(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)))
G(mark(z0), z1) → c6(G(z0, z1))
H(ok(z0)) → c10(H(z0))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

(21) 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(ok(z0)) → c2(TOP(active(z0)))
We considered the (Usable) Rules:

proper(b) → ok(b)
active(a) → mark(b)
proper(a) → ok(a)
And the Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = 0   
POL(G(x1, x2)) = [2]x1   
POL(H(x1)) = 0   
POL(TOP(x1)) = x1   
POL(a) = [1]   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [1] + x1   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(a) → mark(b)
proper(b) → ok(b)
proper(a) → ok(a)
Tuples:

F(mark(z0), z1) → c(F(z0, z1))
F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(mark(z0), z1) → c6(G(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
H(mark(z0)) → c9(H(z0))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
TOP(mark(z0)) → c3(TOP(proper(z0)))
S tuples:none
K tuples:

F(ok(z0), ok(z1)) → c1(F(z0, z1))
G(ok(z0), ok(z1)) → c7(G(z0, z1))
F(mark(z0), z1) → c(F(z0, z1))
H(mark(z0)) → c9(H(z0))
TOP(mark(z0)) → c3(TOP(proper(z0)))
G(mark(z0), z1) → c6(G(z0, z1))
H(ok(z0)) → c10(H(z0))
TOP(ok(z0)) → c2(TOP(active(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, G, H, TOP

Compound Symbols:

c, c1, c6, c7, c9, c10, c2, c3

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

The set S is empty

(24) BOUNDS(1, 1)