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

(0) 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:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)

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:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))

(2) 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:

g(c(h(0))) → g(d(1))
g(d(x)) → x
g(h(x)) → g(x)
g(c(x)) → x
g(c(1)) → g(d(h(0)))

Rewrite Strategy: INNERMOST

(3) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2.
The certificate found is represented by the following graph.

Start state: 1
Accept states: [2]
Transitions:
1→2[g_1|0, h_1|1, c_1|1, 0|1, d_1|1, 1|1, g_1|1, 1|2]
1→3[g_1|1]
1→5[g_1|1]
1→7[h_1|2]
2→2[c_1|0, h_1|0, 0|0, d_1|0, 1|0]
3→4[d_1|1]
4→2[1|1]
5→6[d_1|1]
6→7[h_1|1]
7→2[0|1]

(4) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(h(0))) → g(d(1))
g(d(z0)) → z0
g(h(z0)) → g(z0)
g(c(z0)) → z0
g(c(1)) → g(d(h(0)))

Tuples:

G(c(h(0))) → c1(G(d(1)))
G(d(z0)) → c2
G(h(z0)) → c3(G(z0))
G(c(z0)) → c4
G(c(1)) → c5(G(d(h(0))))

S tuples:

G(c(h(0))) → c1(G(d(1)))
G(d(z0)) → c2
G(h(z0)) → c3(G(z0))
G(c(z0)) → c4
G(c(1)) → c5(G(d(h(0))))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c1, c2, c3, c4, c5

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

Removed 4 trailing nodes:

G(d(z0)) → c2
G(c(1)) → c5(G(d(h(0))))
G(c(z0)) → c4
G(c(h(0))) → c1(G(d(1)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(c(h(0))) → g(d(1))
g(d(z0)) → z0
g(h(z0)) → g(z0)
g(c(z0)) → z0
g(c(1)) → g(d(h(0)))

Tuples:

G(h(z0)) → c3(G(z0))

S tuples:

G(h(z0)) → c3(G(z0))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c3

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(c(h(0))) → g(d(1))
g(d(z0)) → z0
g(h(z0)) → g(z0)
g(c(z0)) → z0
g(c(1)) → g(d(h(0)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(h(z0)) → c3(G(z0))

S tuples:

G(h(z0)) → c3(G(z0))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c3

(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(h(z0)) → c3(G(z0))

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

G(h(z0)) → c3(G(z0))

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

POL(G(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(h(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(h(z0)) → c3(G(z0))

S tuples:none
K tuples:

G(h(z0)) → c3(G(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c3

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

The set S is empty

(14) BOUNDS(1, 1)