(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(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(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:
k(x, h(x), a) → h(x)
k(f(x), y, x) → 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:

h(g(x)) → g(h(f(x)))
f(a) → g(h(a))

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[h_1|0, f_1|0]
1→3[g_1|1]
1→5[g_1|1]
2→2[g_1|0, a|0]
3→4[h_1|1]
3→7[g_1|2]
4→2[f_1|1]
4→5[g_1|1]
5→6[h_1|1]
6→2[a|1]
7→8[h_1|2]
8→5[f_1|2]

(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:

h(g(z0)) → g(h(f(z0)))
f(a) → g(h(a))
Tuples:

H(g(z0)) → c(H(f(z0)), F(z0))
F(a) → c1(H(a))
S tuples:

H(g(z0)) → c(H(f(z0)), F(z0))
F(a) → c1(H(a))
K tuples:none
Defined Rule Symbols:

h, f

Defined Pair Symbols:

H, F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(a) → c1(H(a))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(g(z0)) → g(h(f(z0)))
f(a) → g(h(a))
Tuples:

H(g(z0)) → c(H(f(z0)), F(z0))
S tuples:

H(g(z0)) → c(H(f(z0)), F(z0))
K tuples:none
Defined Rule Symbols:

h, f

Defined Pair Symbols:

H

Compound Symbols:

c

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

h(g(z0)) → g(h(f(z0)))
f(a) → g(h(a))
Tuples:

H(g(z0)) → c(H(f(z0)))
S tuples:

H(g(z0)) → c(H(f(z0)))
K tuples:none
Defined Rule Symbols:

h, f

Defined Pair Symbols:

H

Compound Symbols:

c

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(g(z0)) → g(h(f(z0)))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a))
Tuples:

H(g(z0)) → c(H(f(z0)))
S tuples:

H(g(z0)) → c(H(f(z0)))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

H

Compound Symbols:

c

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

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

H(g(z0)) → c(H(f(z0)))
We considered the (Usable) Rules:

f(a) → g(h(a))
And the Tuples:

H(g(z0)) → c(H(f(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(H(x1)) = x1   
POL(a) = [1]   
POL(c(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = [1] + x1   
POL(h(x1)) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a))
Tuples:

H(g(z0)) → c(H(f(z0)))
S tuples:none
K tuples:

H(g(z0)) → c(H(f(z0)))
Defined Rule Symbols:

f

Defined Pair Symbols:

H

Compound Symbols:

c

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

The set S is empty

(16) BOUNDS(1, 1)