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 (⇔, 147 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)), 81 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 10 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 20 ms)
        ↳18 CdtProblem
        ↳19 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳20 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(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(b, X, c)) → mark(f(X, c, X))
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))

(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:

top(ok(X)) → top(active(X))
proper(b) → ok(b)
proper(c) → ok(c)
active(c) → mark(b)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
top(mark(X)) → top(proper(X))

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]
transitions:
ok0(0) → 0
b0() → 0
c0() → 0
mark0(0) → 0
top0(0) → 1
proper0(0) → 2
active0(0) → 3
f0(0, 0, 0) → 4
active1(0) → 5
top1(5) → 1
b1() → 6
ok1(6) → 2
c1() → 7
ok1(7) → 2
b1() → 8
mark1(8) → 3
f1(0, 0, 0) → 9
ok1(9) → 4
f1(0, 0, 0) → 10
mark1(10) → 4
proper1(0) → 11
top1(11) → 1
ok1(6) → 11
ok1(7) → 11
mark1(8) → 5
ok1(9) → 9
ok1(9) → 10
mark1(10) → 9
mark1(10) → 10
active2(6) → 12
top2(12) → 1
active2(7) → 12
proper2(8) → 13
top2(13) → 1
b2() → 14
ok2(14) → 13
b2() → 15
mark2(15) → 12
active3(14) → 16
top3(16) → 1
proper3(15) → 17
top3(17) → 1
b3() → 18
ok3(18) → 17
active4(18) → 19
top4(19) → 1

(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:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(c) → ok(c)
active(c) → mark(b)
f(ok(z0), ok(z1), ok(z2)) → ok(f(z0, z1, z2))
f(z0, mark(z1), z2) → mark(f(z0, z1, z2))
Tuples:

TOP(ok(z0)) → c1(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c3
PROPER(c) → c4
ACTIVE(c) → c5
F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
S tuples:

TOP(ok(z0)) → c1(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c3
PROPER(c) → c4
ACTIVE(c) → c5
F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

top, proper, active, f

Defined Pair Symbols:

TOP, PROPER, ACTIVE, F

Compound Symbols:

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

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

Removed 3 trailing nodes:

PROPER(c) → c4
PROPER(b) → c3
ACTIVE(c) → c5

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(c) → ok(c)
active(c) → mark(b)
f(ok(z0), ok(z1), ok(z2)) → ok(f(z0, z1, z2))
f(z0, mark(z1), z2) → mark(f(z0, z1, z2))
Tuples:

TOP(ok(z0)) → c1(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
S tuples:

TOP(ok(z0)) → c1(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c2(TOP(proper(z0)), PROPER(z0))
F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

top, proper, active, f

Defined Pair Symbols:

TOP, F

Compound Symbols:

c1, c2, c6, c7

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

Removed 2 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(b) → ok(b)
proper(c) → ok(c)
active(c) → mark(b)
f(ok(z0), ok(z1), ok(z2)) → ok(f(z0, z1, z2))
f(z0, mark(z1), z2) → mark(f(z0, z1, z2))
Tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

top, proper, active, f

Defined Pair Symbols:

F, TOP

Compound Symbols:

c6, c7, c1, c2

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
f(ok(z0), ok(z1), ok(z2)) → ok(f(z0, z1, z2))
f(z0, mark(z1), z2) → mark(f(z0, z1, z2))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c6, c7, c1, c2

(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), ok(z2)) → c6(F(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = [2]x3   
POL(TOP(x1)) = 0   
POL(active(x1)) = 0   
POL(b) = [3]   
POL(c) = [2]   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(mark(x1)) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [2]x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c6, c7, c1, 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.

F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
We considered the (Usable) Rules:

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = x2   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c) = [1]   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [1] + x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:

TOP(mark(z0)) → c2(TOP(proper(z0)))
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c6, c7, c1, c2

(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.

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

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = x3   
POL(TOP(x1)) = x1   
POL(active(x1)) = [2] + x1   
POL(b) = [1]   
POL(c) = [2]   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(mark(x1)) = [3] + x1   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = [2] + x1   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
S tuples:none
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c6(F(z0, z1, z2))
F(z0, mark(z1), z2) → c7(F(z0, z1, z2))
TOP(ok(z0)) → c1(TOP(active(z0)))
TOP(mark(z0)) → c2(TOP(proper(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c6, c7, c1, c2

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

The set S is empty

(20) BOUNDS(1, 1)