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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 123 ms)
↳6 CdtProblem
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (⇔, 0 ms)
↳10 BOUNDS(O(1), O(1))

(0) Obligation:

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

g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(z0, s(z1))) → g(c(s(z0), z1))
f(c(s(z0), z1)) → f(c(z0, s(z1)))
f(f(z0)) → f(d(f(z0)))
f(z0) → z0
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
F(f(z0)) → c3(F(d(f(z0))), F(z0))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
F(f(z0)) → c3(F(d(f(z0))), F(z0))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c2, c3

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

F(f(z0)) → c3(F(d(f(z0))), F(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(z0, s(z1))) → g(c(s(z0), z1))
f(c(s(z0), z1)) → f(c(z0, s(z1)))
f(f(z0)) → f(d(f(z0)))
f(z0) → z0
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c2

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
We considered the (Usable) Rules:none
And the Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = [4]x1   
POL(G(x1)) = 0   
POL(c(x1, x2)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [2] + x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(z0, s(z1))) → g(c(s(z0), z1))
f(c(s(z0), z1)) → f(c(z0, s(z1)))
f(f(z0)) → f(d(f(z0)))
f(z0) → z0
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
K tuples:

F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c2

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
We considered the (Usable) Rules:none
And the Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = 0   
POL(G(x1)) = [4]x1   
POL(c(x1, x2)) = x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [4] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(z0, s(z1))) → g(c(s(z0), z1))
f(c(s(z0), z1)) → f(c(z0, s(z1)))
f(f(z0)) → f(d(f(z0)))
f(z0) → z0
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:none
K tuples:

F(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c2

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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