(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) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)

Rewrite Strategy: INNERMOST

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

The following defined symbols can occur below the 0th argument of g: f, g
The following defined symbols can occur below the 1th argument of g: f, g

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(g(x, y)) → g(f(x), f(y))

(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:

f(a) → b
f(c) → d
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(a) → c1
F(c) → c2
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c4
S tuples:

F(a) → c1
F(c) → c2
F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
G(z0, z0) → c4
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2, c3, c4

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

Removed 3 trailing nodes:

G(z0, z0) → c4
F(a) → c1
F(c) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
S tuples:

F(h(z0, z1)) → c3(G(h(z1, f(z0)), h(z0, f(z1))), F(z0), F(z1))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c3

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)
Tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c3

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(a) → b
f(c) → d
f(h(z0, z1)) → g(h(z1, f(z0)), h(z0, f(z1)))
g(z0, z0) → h(e, z0)

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

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.

F(h(z0, z1)) → c3(F(z0), F(z1))
We considered the (Usable) Rules:none
And the Tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = [2] + [2]x1   
POL(c3(x1, x2)) = x1 + x2   
POL(h(x1, x2)) = [2] + x1 + x2   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
S tuples:none
K tuples:

F(h(z0, z1)) → c3(F(z0), F(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3

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

The set S is empty

(14) BOUNDS(1, 1)