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

(0) Obligation:

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


The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Rewrite Strategy: FULL

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

The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(X, Y, g(X, Y)) → h(0, g(X, Y))

(2) Obligation:

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


The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
g(X, s(Y)) → g(X, Y)
g(0, Y) → 0

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

(4) Obligation:

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


The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
g(X, s(Y)) → g(X, Y)
g(0, Y) → 0

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(z0, z1) → f(z0, s(z0), z1)
g(z0, s(z1)) → g(z0, z1)
g(0, z0) → 0

Tuples:

H(z0, z1) → c
G(z0, s(z1)) → c1(G(z0, z1))
G(0, z0) → c2

S tuples:

H(z0, z1) → c
G(z0, s(z1)) → c1(G(z0, z1))
G(0, z0) → c2

K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

G(0, z0) → c2
H(z0, z1) → c

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(z0, z1) → f(z0, s(z0), z1)
g(z0, s(z1)) → g(z0, z1)
g(0, z0) → 0

Tuples:

G(z0, s(z1)) → c1(G(z0, z1))

S tuples:

G(z0, s(z1)) → c1(G(z0, z1))

K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

G

Compound Symbols:

c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(z0, z1) → f(z0, s(z0), z1)
g(z0, s(z1)) → g(z0, z1)
g(0, z0) → 0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(z0, s(z1)) → c1(G(z0, z1))

S tuples:

G(z0, s(z1)) → c1(G(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

G(z0, s(z1)) → c1(G(z0, z1))

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

G(z0, s(z1)) → c1(G(z0, z1))

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

POL(G(x₁, x₂)) = x₂   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(z0, s(z1)) → c1(G(z0, z1))

S tuples:none
K tuples:

G(z0, s(z1)) → c1(G(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

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

The set S is empty

(14) BOUNDS(1, 1)