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:

eq0(S(x'), S(x)) → eq0(x', x)
eq0(S(x), 0) → 0
eq0(0, S(x)) → 0
eq0(0, 0) → S(0)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

eq0(S(z0), S(z1)) → eq0(z0, z1)
eq0(S(z0), 0) → 0
eq0(0, S(z0)) → 0
eq0(0, 0) → S(0)

Tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
EQ0(S(z0), 0) → c1
EQ0(0, S(z0)) → c2
EQ0(0, 0) → c3

S tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))
EQ0(S(z0), 0) → c1
EQ0(0, S(z0)) → c2
EQ0(0, 0) → c3

K tuples:none
Defined Rule Symbols:

eq0

Defined Pair Symbols:

EQ0

Compound Symbols:

c, c1, c2, c3

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

Removed 3 trailing nodes:

EQ0(0, 0) → c3
EQ0(0, S(z0)) → c2
EQ0(S(z0), 0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

eq0(S(z0), S(z1)) → eq0(z0, z1)
eq0(S(z0), 0) → 0
eq0(0, S(z0)) → 0
eq0(0, 0) → S(0)

Tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

S tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

K tuples:none
Defined Rule Symbols:

eq0

Defined Pair Symbols:

EQ0

Compound Symbols:

c

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

eq0(S(z0), S(z1)) → eq0(z0, z1)
eq0(S(z0), 0) → 0
eq0(0, S(z0)) → 0
eq0(0, 0) → S(0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

S tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

EQ0

Compound Symbols:

c

(7) 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.

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

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

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

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

POL(EQ0(x₁, x₂)) = [2]x₁   
POL(S(x₁)) = [1] + x₁   
POL(c(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

S tuples:none
K tuples:

EQ0(S(z0), S(z1)) → c(EQ0(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

EQ0

Compound Symbols:

c

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

The set S is empty

(10) BOUNDS(O(1), O(1))