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:

f(x, x) → a
f(g(x), y) → f(x, y)

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

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

(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(x, x) → a
f(g(x), y) → f(x, y)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, z0) → a
f(g(z0), z1) → f(z0, z1)

Tuples:

F(z0, z0) → c
F(g(z0), z1) → c1(F(z0, z1))

S tuples:

F(z0, z0) → c
F(g(z0), z1) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(z0, z0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, z0) → a
f(g(z0), z1) → f(z0, z1)

Tuples:

F(g(z0), z1) → c1(F(z0, z1))

S tuples:

F(g(z0), z1) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, z0) → a
f(g(z0), z1) → f(z0, z1)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1) → c1(F(z0, z1))

S tuples:

F(g(z0), z1) → c1(F(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

(9) 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(g(z0), z1) → c1(F(z0, z1))

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

F(g(z0), z1) → c1(F(z0, z1))

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

POL(F(x₁, x₂)) = x₁   
POL(c1(x₁)) = x₁   
POL(g(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1) → c1(F(z0, z1))

S tuples:none
K tuples:

F(g(z0), z1) → c1(F(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty

(12) BOUNDS(1, 1)