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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 13 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 104 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 23 ms)
        ↳12 CdtProblem
        ↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 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:

plus#2(0, x12) → x12
plus#2(S(x4), x2) → S(plus#2(x4, x2))
fold#3(Nil) → 0
fold#3(Cons(x4, x2)) → plus#2(x4, fold#3(x2))
main(x1) → fold#3(x1)

Rewrite Strategy: INNERMOST

(1) 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 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
00() → 0
S0(0) → 0
Nil0() → 0
Cons0(0, 0) → 0
plus#20(0, 0) → 1
fold#30(0) → 2
main0(0) → 3
plus#21(0, 0) → 4
S1(4) → 1
01() → 2
fold#31(0) → 5
plus#21(0, 5) → 2
fold#31(0) → 3
plus#21(0, 5) → 4
S1(4) → 2
S1(4) → 4
01() → 3
01() → 5
plus#21(0, 5) → 3
plus#21(0, 5) → 5
S1(4) → 3
S1(4) → 5
0 → 1
0 → 4
5 → 2
5 → 3
5 → 4

(2) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus#2(0, z0) → z0
plus#2(S(z0), z1) → S(plus#2(z0, z1))
fold#3(Nil) → 0
fold#3(Cons(z0, z1)) → plus#2(z0, fold#3(z1))
main(z0) → fold#3(z0)
Tuples:

PLUS#2(0, z0) → c
PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Nil) → c2
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
MAIN(z0) → c4(FOLD#3(z0))
S tuples:

PLUS#2(0, z0) → c
PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Nil) → c2
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
MAIN(z0) → c4(FOLD#3(z0))
K tuples:none
Defined Rule Symbols:

plus#2, fold#3, main

Defined Pair Symbols:

PLUS#2, FOLD#3, MAIN

Compound Symbols:

c, c1, c2, c3, c4

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

MAIN(z0) → c4(FOLD#3(z0))
Removed 2 trailing nodes:

PLUS#2(0, z0) → c
FOLD#3(Nil) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus#2(0, z0) → z0
plus#2(S(z0), z1) → S(plus#2(z0, z1))
fold#3(Nil) → 0
fold#3(Cons(z0, z1)) → plus#2(z0, fold#3(z1))
main(z0) → fold#3(z0)
Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
S tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
K tuples:none
Defined Rule Symbols:

plus#2, fold#3, main

Defined Pair Symbols:

PLUS#2, FOLD#3

Compound Symbols:

c1, c3

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

main(z0) → fold#3(z0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

fold#3(Nil) → 0
fold#3(Cons(z0, z1)) → plus#2(z0, fold#3(z1))
plus#2(0, z0) → z0
plus#2(S(z0), z1) → S(plus#2(z0, z1))
Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
S tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
K tuples:none
Defined Rule Symbols:

fold#3, plus#2

Defined Pair Symbols:

PLUS#2, FOLD#3

Compound Symbols:

c1, c3

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

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

FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(Cons(x1, x2)) = [2] + x1 + x2   
POL(FOLD#3(x1)) = [2]x12   
POL(Nil) = [2]   
POL(PLUS#2(x1, x2)) = [2]   
POL(S(x1)) = [2]   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(fold#3(x1)) = [2] + x1 + x12   
POL(plus#2(x1, x2)) = [1] + [2]x1 + x2 + x12   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

fold#3(Nil) → 0
fold#3(Cons(z0, z1)) → plus#2(z0, fold#3(z1))
plus#2(0, z0) → z0
plus#2(S(z0), z1) → S(plus#2(z0, z1))
Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
S tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
K tuples:

FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
Defined Rule Symbols:

fold#3, plus#2

Defined Pair Symbols:

PLUS#2, FOLD#3

Compound Symbols:

c1, c3

(11) 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#2(S(z0), z1) → c1(PLUS#2(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(Cons(x1, x2)) = [1] + x1 + x2   
POL(FOLD#3(x1)) = x1   
POL(Nil) = 0   
POL(PLUS#2(x1, x2)) = x1   
POL(S(x1)) = [1] + x1   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(fold#3(x1)) = 0   
POL(plus#2(x1, x2)) = 0   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

fold#3(Nil) → 0
fold#3(Cons(z0, z1)) → plus#2(z0, fold#3(z1))
plus#2(0, z0) → z0
plus#2(S(z0), z1) → S(plus#2(z0, z1))
Tuples:

PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
S tuples:none
K tuples:

FOLD#3(Cons(z0, z1)) → c3(PLUS#2(z0, fold#3(z1)), FOLD#3(z1))
PLUS#2(S(z0), z1) → c1(PLUS#2(z0, z1))
Defined Rule Symbols:

fold#3, plus#2

Defined Pair Symbols:

PLUS#2, FOLD#3

Compound Symbols:

c1, c3

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

The set S is empty

(14) BOUNDS(1, 1)