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 (⇔, 145 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 3 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)), 82 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 24 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 25 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, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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: active, proper
The following defined symbols can occur below the 0th argument of active: active, proper
The following defined symbols can occur below the 0th argument of proper: active, proper

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

(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:

active(d) → m(b)
top(ok(x)) → top(active(x))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
f(x, y, mark(z)) → mark(f(x, y, z))
proper(d) → ok(d)
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:
d0() → 0
m0(0) → 0
b0() → 0
ok0(0) → 0
mark0(0) → 0
c0() → 0
active0(0) → 1
top0(0) → 2
proper0(0) → 3
f0(0, 0, 0) → 4
b1() → 5
m1(5) → 1
active1(0) → 6
top1(6) → 2
c1() → 7
mark1(7) → 1
b1() → 8
ok1(8) → 3
c1() → 9
ok1(9) → 3
f1(0, 0, 0) → 10
mark1(10) → 4
d1() → 11
ok1(11) → 3
f1(0, 0, 0) → 12
ok1(12) → 4
proper1(0) → 13
top1(13) → 2
m1(5) → 6
mark1(7) → 6
ok1(8) → 13
ok1(9) → 13
mark1(10) → 10
mark1(10) → 12
ok1(11) → 13
ok1(12) → 10
ok1(12) → 12
active2(8) → 14
top2(14) → 2
active2(9) → 14
active2(11) → 14
proper2(7) → 15
top2(15) → 2
b2() → 16
m2(16) → 14
c2() → 17
mark2(17) → 14
c2() → 18
ok2(18) → 15
active3(18) → 19
top3(19) → 2
proper3(17) → 20
top3(20) → 2
c3() → 21
ok3(21) → 20
active4(21) → 22
top4(22) → 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:

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

ACTIVE(d) → c1
ACTIVE(d) → c2
TOP(ok(z0)) → c3(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c4(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c5
PROPER(c) → c6
PROPER(d) → c7
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
S tuples:

ACTIVE(d) → c1
ACTIVE(d) → c2
TOP(ok(z0)) → c3(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c4(TOP(proper(z0)), PROPER(z0))
PROPER(b) → c5
PROPER(c) → c6
PROPER(d) → c7
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

active, top, proper, f

Defined Pair Symbols:

ACTIVE, TOP, PROPER, F

Compound Symbols:

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

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

Removed 5 trailing nodes:

PROPER(c) → c6
PROPER(b) → c5
ACTIVE(d) → c1
ACTIVE(d) → c2
PROPER(d) → c7

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

TOP(ok(z0)) → c3(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c4(TOP(proper(z0)), PROPER(z0))
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
S tuples:

TOP(ok(z0)) → c3(TOP(active(z0)), ACTIVE(z0))
TOP(mark(z0)) → c4(TOP(proper(z0)), PROPER(z0))
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
K tuples:none
Defined Rule Symbols:

active, top, proper, f

Defined Pair Symbols:

TOP, F

Compound Symbols:

c3, c4, c8, c9

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

Removed 2 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
S tuples:

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

active, top, proper, f

Defined Pair Symbols:

F, TOP

Compound Symbols:

c8, c9, c3, c4

(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(z0, z1, mark(z2)) → mark(f(z0, z1, z2))
f(ok(z0), ok(z1), ok(z2)) → ok(f(z0, z1, z2))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
S tuples:

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c8, c9, c3, c4

(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)) → c9(F(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = [2]x1   
POL(TOP(x1)) = 0   
POL(active(x1)) = [2] + [3]x1   
POL(b) = [2]   
POL(c) = [2]   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(d) = [2]   
POL(m(x1)) = [3]   
POL(mark(x1)) = [3]   
POL(ok(x1)) = [3] + x1   
POL(proper(x1)) = [2] + [3]x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
S tuples:

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c8, c9, c3, c4

(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, z1, mark(z2)) → c8(F(z0, z1, z2))
TOP(mark(z0)) → c4(TOP(proper(z0)))
We considered the (Usable) Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(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)) = x1   
POL(b) = 0   
POL(c) = 0   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(d) = [1]   
POL(m(x1)) = 0   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
S tuples:

TOP(ok(z0)) → c3(TOP(active(z0)))
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
TOP(mark(z0)) → c4(TOP(proper(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c8, c9, c3, c4

(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(ok(z0)) → c3(TOP(active(z0)))
We considered the (Usable) Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2, x3)) = x1   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(b) = 0   
POL(c) = 0   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(d) = [1]   
POL(m(x1)) = 0   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [1] + x1   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
TOP(ok(z0)) → c3(TOP(active(z0)))
TOP(mark(z0)) → c4(TOP(proper(z0)))
S tuples:none
K tuples:

F(ok(z0), ok(z1), ok(z2)) → c9(F(z0, z1, z2))
F(z0, z1, mark(z2)) → c8(F(z0, z1, z2))
TOP(mark(z0)) → c4(TOP(proper(z0)))
TOP(ok(z0)) → c3(TOP(active(z0)))
Defined Rule Symbols:

active, proper

Defined Pair Symbols:

F, TOP

Compound Symbols:

c8, c9, c3, c4

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

The set S is empty

(20) BOUNDS(1, 1)