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(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))

Rewrite Strategy: FULL

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

The following defined symbols can occur below the 1th argument of h: d
The following defined symbols can occur below the 1th argument of d: d

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, 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(z, e(x)) → h(c(z), d(z, x))
g(e(x), e(y)) → e(g(x, y))

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(z, e(x)) → h(c(z), d(z, x))
g(e(x), e(y)) → e(g(x, y))

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, e(z1)) → h(c(z0), d(z0, z1))
g(e(z0), e(z1)) → e(g(z0, z1))

Tuples:

H(z0, e(z1)) → c1(H(c(z0), d(z0, z1)))
G(e(z0), e(z1)) → c2(G(z0, z1))

S tuples:

H(z0, e(z1)) → c1(H(c(z0), d(z0, z1)))
G(e(z0), e(z1)) → c2(G(z0, z1))

K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

H, G

Compound Symbols:

c1, c2

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

Removed 1 trailing nodes:

H(z0, e(z1)) → c1(H(c(z0), d(z0, z1)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(z0, e(z1)) → h(c(z0), d(z0, z1))
g(e(z0), e(z1)) → e(g(z0, z1))

Tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

S tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

G

Compound Symbols:

c2

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(z0, e(z1)) → h(c(z0), d(z0, z1))
g(e(z0), e(z1)) → e(g(z0, z1))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

S tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c2

(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(e(z0), e(z1)) → c2(G(z0, z1))

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

G(e(z0), e(z1)) → c2(G(z0, z1))

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

POL(G(x₁, x₂)) = x₂   
POL(c2(x₁)) = x₁   
POL(e(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

S tuples:none
K tuples:

G(e(z0), e(z1)) → c2(G(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c2

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

The set S is empty

(14) BOUNDS(1, 1)