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 (⇔, 7 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 13 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)), 10 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 19 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(s(X)) → s(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:

top(ok(X)) → top(active(X))
proper(tt) → ok(tt)
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
proper(0) → ok(0)
and(mark(X1), X2) → mark(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
top(mark(X)) → top(proper(X))
and(ok(X1), ok(X2)) → ok(and(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 2.

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]
transitions:
ok0(0) → 0
active0(0) → 0
tt0() → 0
mark0(0) → 0
00() → 0
top0(0) → 1
proper0(0) → 2
s0(0) → 3
plus0(0, 0) → 4
and0(0, 0) → 5
active1(0) → 6
top1(6) → 1
tt1() → 7
ok1(7) → 2
s1(0) → 8
ok1(8) → 3
s1(0) → 9
mark1(9) → 3
plus1(0, 0) → 10
mark1(10) → 4
01() → 11
ok1(11) → 2
and1(0, 0) → 12
mark1(12) → 5
plus1(0, 0) → 13
ok1(13) → 4
proper1(0) → 14
top1(14) → 1
and1(0, 0) → 15
ok1(15) → 5
ok1(7) → 14
ok1(8) → 8
ok1(8) → 9
mark1(9) → 8
mark1(9) → 9
mark1(10) → 10
mark1(10) → 13
ok1(11) → 14
mark1(12) → 12
mark1(12) → 15
ok1(13) → 10
ok1(13) → 13
ok1(15) → 12
ok1(15) → 15
active2(7) → 16
top2(16) → 1
active2(11) → 16

(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(tt) → ok(tt)
proper(0) → ok(0)
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
plus(mark(z0), z1) → mark(plus(z0, z1))
plus(ok(z0), ok(z1)) → ok(plus(z0, z1))
plus(z0, mark(z1)) → mark(plus(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
Tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
PROPER(tt) → c2
PROPER(0) → c3
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
S tuples:

TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
PROPER(tt) → c2
PROPER(0) → c3
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
K tuples:none
Defined Rule Symbols:

top, proper, s, plus, and

Defined Pair Symbols:

TOP, PROPER, S, PLUS, AND

Compound Symbols:

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

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

Removed 3 trailing nodes:

PROPER(0) → c3
PROPER(tt) → c2
TOP(ok(z0)) → c(TOP(active(z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(tt) → ok(tt)
proper(0) → ok(0)
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
plus(mark(z0), z1) → mark(plus(z0, z1))
plus(ok(z0), ok(z1)) → ok(plus(z0, z1))
plus(z0, mark(z1)) → mark(plus(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
Tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
S tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)), PROPER(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
K tuples:none
Defined Rule Symbols:

top, proper, s, plus, and

Defined Pair Symbols:

TOP, S, PLUS, AND

Compound Symbols:

c1, c4, c5, c6, c7, c8, c9, c10

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
proper(tt) → ok(tt)
proper(0) → ok(0)
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
plus(mark(z0), z1) → mark(plus(z0, z1))
plus(ok(z0), ok(z1)) → ok(plus(z0, z1))
plus(z0, mark(z1)) → mark(plus(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))
Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

top, proper, s, plus, and

Defined Pair Symbols:

S, PLUS, AND, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c1

(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))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
plus(mark(z0), z1) → mark(plus(z0, z1))
plus(ok(z0), ok(z1)) → ok(plus(z0, z1))
plus(z0, mark(z1)) → mark(plus(z0, z1))
and(mark(z0), z1) → mark(and(z0, z1))
and(ok(z0), ok(z1)) → ok(and(z0, z1))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(tt) → ok(tt)
proper(0) → ok(0)
Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

S, PLUS, AND, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c1

(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.

TOP(mark(z0)) → c1(TOP(proper(z0)))
We considered the (Usable) Rules:

proper(tt) → ok(tt)
proper(0) → ok(0)
And the Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(AND(x1, x2)) = 0   
POL(PLUS(x1, x2)) = 0   
POL(S(x1)) = 0   
POL(TOP(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1]   
POL(ok(x1)) = 0   
POL(proper(x1)) = 0   
POL(tt) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(tt) → ok(tt)
proper(0) → ok(0)
Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

S, PLUS, AND, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c1

(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.

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(AND(x1, x2)) = 0   
POL(PLUS(x1, x2)) = [2]x2   
POL(S(x1)) = [2]x1   
POL(TOP(x1)) = 0   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [2] + x1   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = 0   
POL(tt) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(tt) → ok(tt)
proper(0) → ok(0)
Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
S tuples:

PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

S, PLUS, AND, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c1

(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.

PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(AND(x1, x2)) = x1   
POL(PLUS(x1, x2)) = x1   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = [1] + [3]x1   
POL(tt) = [1]   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(tt) → ok(tt)
proper(0) → ok(0)
Tuples:

S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
TOP(mark(z0)) → c1(TOP(proper(z0)))
S tuples:none
K tuples:

TOP(mark(z0)) → c1(TOP(proper(z0)))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
PLUS(ok(z0), ok(z1)) → c7(PLUS(z0, z1))
PLUS(z0, mark(z1)) → c8(PLUS(z0, z1))
PLUS(mark(z0), z1) → c6(PLUS(z0, z1))
AND(mark(z0), z1) → c9(AND(z0, z1))
AND(ok(z0), ok(z1)) → c10(AND(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

S, PLUS, AND, TOP

Compound Symbols:

c4, c5, c6, c7, c8, c9, c10, c1

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

The set S is empty

(20) BOUNDS(1, 1)