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

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

dbl(S(0), S(0)) → S(S(S(S(0))))
save(S(x)) → dbl(0, save(x))
save(0) → 0
dbl(0, y) → y

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0))))
dbl(0, z0) → z0
save(S(z0)) → dbl(0, save(z0))
save(0) → 0

Tuples:

DBL(S(0), S(0)) → c
DBL(0, z0) → c1
SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))
SAVE(0) → c3

S tuples:

DBL(S(0), S(0)) → c
DBL(0, z0) → c1
SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))
SAVE(0) → c3

K tuples:none
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

DBL, SAVE

Compound Symbols:

c, c1, c2, c3

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

Removed 3 trailing nodes:

SAVE(0) → c3
DBL(0, z0) → c1
DBL(S(0), S(0)) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0))))
dbl(0, z0) → z0
save(S(z0)) → dbl(0, save(z0))
save(0) → 0

Tuples:

SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))

S tuples:

SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))

K tuples:none
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0))))
dbl(0, z0) → z0
save(S(z0)) → dbl(0, save(z0))
save(0) → 0

Tuples:

SAVE(S(z0)) → c2(SAVE(z0))

S tuples:

SAVE(S(z0)) → c2(SAVE(z0))

K tuples:none
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

dbl(S(0), S(0)) → S(S(S(S(0))))
dbl(0, z0) → z0
save(S(z0)) → dbl(0, save(z0))
save(0) → 0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SAVE(S(z0)) → c2(SAVE(z0))

S tuples:

SAVE(S(z0)) → c2(SAVE(z0))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

(9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

SAVE(S(z0)) → c2(SAVE(z0))

We considered the (Usable) Rules:none
And the Tuples:

SAVE(S(z0)) → c2(SAVE(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(S(x₁)) = [1] + x₁   
POL(SAVE(x₁)) = [5]x₁   
POL(c2(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SAVE(S(z0)) → c2(SAVE(z0))

S tuples:none
K tuples:

SAVE(S(z0)) → c2(SAVE(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

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

The set S is empty

(12) BOUNDS(O(1), O(1))