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

0 CpxTRS
↳1 DependencyGraphProof (BOTH BOUNDS(ID, ID), 14 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsProof (⇔, 0 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 CdtUsableRulesProof (⇔, 0 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:

f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0

Rewrite Strategy: INNERMOST

(1) DependencyGraphProof (BOTH BOUNDS(ID, ID) transformation)

The following rules are not reachable from basic terms in the dependency graph and can be removed:
f(g(x), s(0), y) → f(y, y, g(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:

g(s(x)) → s(g(x))
g(0) → 0

Rewrite Strategy: INNERMOST

(3) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[g_1|0, 0|1]
1→3[s_1|1]
2→2[s_1|0, 0|0]
3→2[g_1|1, 0|1]
3→3[s_1|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:

g(s(z0)) → s(g(z0))
g(0) → 0
Tuples:

G(s(z0)) → c(G(z0))
G(0) → c1
S tuples:

G(s(z0)) → c(G(z0))
G(0) → c1
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

G(0) → c1

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(s(z0)) → s(g(z0))
g(0) → 0
Tuples:

G(s(z0)) → c(G(z0))
S tuples:

G(s(z0)) → c(G(z0))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(s(z0)) → s(g(z0))
g(0) → 0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c(G(z0))
S tuples:

G(s(z0)) → c(G(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c

(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.

G(s(z0)) → c(G(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0)) → c(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1)) = x1   
POL(c(x1)) = x1   
POL(s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c(G(z0))
S tuples:none
K tuples:

G(s(z0)) → c(G(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c

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

The set S is empty

(14) BOUNDS(1, 1)