The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

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

(0) Obligation:

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
i(f(z0, h(z1))) → z1
i(h2(s(z0), z1, h1(z0, z2))) → z2
k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
Tuples:

F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
S tuples:

F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
K tuples:none
Defined Rule Symbols:

f, g, h2, i, k

Defined Pair Symbols:

F, G, H2

Compound Symbols:

c, c1, c2, c3

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
i(f(z0, h(z1))) → z1
i(h2(s(z0), z1, h1(z0, z2))) → z2
k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
K tuples:none
Defined Rule Symbols:

f, g, h2, i, k

Defined Pair Symbols:

F, G, H2

Compound Symbols:

c1, c2, c3, c

(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(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
We considered the (Usable) Rules:

k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
And the Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(F(x1, x2)) = [5]x1 + [2]x2   
POL(G(x1)) = [3]   
POL(H2(x1, x2, x3)) = [1] + [5]x1   
POL(c(x1, x2)) = x1 + x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(f(x1, x2)) = [2] + [2]x1 + [3]x2   
POL(g(x1)) = [2]   
POL(h(x1)) = [4] + x1   
POL(h1(x1, x2)) = [2] + x1   
POL(h2(x1, x2, x3)) = [4]x1 + x2   
POL(j(x1, x2)) = [1] + x1 + x2   
POL(k(x1)) = [3]x1   
POL(s(x1)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
i(f(z0, h(z1))) → z1
i(h2(s(z0), z1, h1(z0, z2))) → z2
k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
K tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
Defined Rule Symbols:

f, g, h2, i, k

Defined Pair Symbols:

F, G, H2

Compound Symbols:

c1, c2, c3, c

(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.

F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
We considered the (Usable) Rules:

k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
And the Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(F(x1, x2)) = [3] + [3]x1 + x2   
POL(G(x1)) = [4]   
POL(H2(x1, x2, x3)) = [1] + x1 + x3   
POL(c(x1, x2)) = x1 + x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(f(x1, x2)) = [5] + [3]x1 + [4]x2   
POL(g(x1)) = [2]   
POL(h(x1)) = [3] + x1   
POL(h1(x1, x2)) = [2]   
POL(h2(x1, x2, x3)) = [3]x1 + [5]x2   
POL(j(x1, x2)) = [2] + x1   
POL(k(x1)) = x1   
POL(s(x1)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
i(f(z0, h(z1))) → z1
i(h2(s(z0), z1, h1(z0, z2))) → z2
k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
K tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
Defined Rule Symbols:

f, g, h2, i, k

Defined Pair Symbols:

F, G, H2

Compound Symbols:

c1, c2, c3, c

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
We considered the (Usable) Rules:

k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
And the Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(F(x1, x2)) = [2] + x1 + [2]x12   
POL(G(x1)) = [1] + [2]x1   
POL(H2(x1, x2, x3)) = x2 + x1·x2   
POL(c(x1, x2)) = x1 + x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(f(x1, x2)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = 0   
POL(h1(x1, x2)) = 0   
POL(h2(x1, x2, x3)) = x2 + [2]x1·x2   
POL(j(x1, x2)) = [2] + x1   
POL(k(x1)) = 0   
POL(s(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(j(z0, z1), z1) → g(f(z0, k(z1)))
f(z0, h1(z1, z2)) → h2(0, z0, h1(z1, z2))
g(h2(z0, z1, h1(z2, z3))) → h2(s(z0), z1, h1(z2, z3))
h2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → h2(s(z0), z1, h1(s(z2), z3))
i(f(z0, h(z1))) → z1
i(h2(s(z0), z1, h1(z0, z2))) → z2
k(h(z0)) → h1(0, z0)
k(h1(z0, z1)) → h1(s(z0), z1)
Tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:none
K tuples:

F(z0, h1(z1, z2)) → c1(H2(0, z0, h1(z1, z2)))
G(h2(z0, z1, h1(z2, z3))) → c2(H2(s(z0), z1, h1(z2, z3)))
F(j(z0, z1), z1) → c(G(f(z0, k(z1))), F(z0, k(z1)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c3(H2(s(z0), z1, h1(s(z2), z3)))
Defined Rule Symbols:

f, g, h2, i, k

Defined Pair Symbols:

F, G, H2

Compound Symbols:

c1, c2, c3, c

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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