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

0 CpxTRS
↳1 DependencyGraphProof (BOTH BOUNDS(ID, ID), 14 ms)
↳2 CpxTRS
↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 140 ms)
↳10 CdtProblem
↳11 CdtKnowledgeProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 829 ms)
↳14 CdtProblem
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 BOUNDS(1, 1)

(0) Obligation:

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


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

The following rules are not reachable from basic terms in the dependency graph and can be removed:
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z

(2) Obligation:

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


The TRS R consists of the following rules:

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

K(h(z0)) → c
K(h1(z0, z1)) → c1
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
S tuples:

K(h(z0)) → c
K(h1(z0, z1)) → c1
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
K tuples:none
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

K, H2, G, F

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 2 trailing nodes:

K(h1(z0, z1)) → c1
K(h(z0)) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)), K(z1))
K tuples:none
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

H2, G, F

Compound Symbols:

c2, c3, c4, c5

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
K tuples:none
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

H2, G, F

Compound Symbols:

c2, c3, c4, c5

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(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)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(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)
And the Tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(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 + x2   
POL(G(x1)) = 0   
POL(H2(x1, x2, x3)) = 0   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1, x2)) = x1 + x2   
POL(f(x1, x2)) = [2] + [3]x1 + [2]x2   
POL(g(x1)) = x1   
POL(h(x1)) = [2] + x1   
POL(h1(x1, x2)) = [3] + x1 + x2   
POL(h2(x1, x2, x3)) = [3]x2 + x3   
POL(j(x1, x2)) = [3] + x1 + x2   
POL(k(x1)) = [1] + [2]x1   
POL(s(x1)) = [3]   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
K tuples:

F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

H2, G, F

Compound Symbols:

c2, c3, c4, c5

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

The following tuples could be moved from S to K by knowledge propagation:

G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
K tuples:

F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

H2, G, F

Compound Symbols:

c2, c3, c4, c5

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(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)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
We considered the (Usable) Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(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)) = [2] + [2]x1 + [2]x1·x2 + [2]x12   
POL(G(x1)) = [1] + [2]x1   
POL(H2(x1, x2, x3)) = [1] + [2]x2 + [2]x3   
POL(c2(x1)) = x1   
POL(c3(x1)) = x1   
POL(c4(x1)) = x1   
POL(c5(x1, x2)) = x1 + x2   
POL(f(x1, x2)) = [1] + [2]x1   
POL(g(x1)) = [2] + x1   
POL(h(x1)) = 0   
POL(h1(x1, x2)) = 0   
POL(h2(x1, x2, x3)) = x1 + x2 + [2]x3 + x1·x2   
POL(j(x1, x2)) = [2] + x1   
POL(k(x1)) = [2]   
POL(s(x1)) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
S tuples:none
K tuples:

F(z0, h1(z1, z2)) → c4(H2(0, z0, h1(z1, z2)))
F(j(z0, z1), z1) → c5(G(f(z0, k(z1))), F(z0, k(z1)))
G(h2(z0, z1, h1(z2, z3))) → c3(H2(s(z0), z1, h1(z2, z3)))
H2(z0, j(z1, h1(z2, z3)), h1(z2, z3)) → c2(H2(s(z0), z1, h1(s(z2), z3)))
Defined Rule Symbols:

k, h2, g, f

Defined Pair Symbols:

H2, G, F

Compound Symbols:

c2, c3, c4, c5

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

The set S is empty

(16) BOUNDS(1, 1)