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(0, y) → 0
f(s(x), y) → f(f(x, y), y)

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
f(s(x), [])

The defined contexts are:
f([], x1)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, 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(0, y) → 0
f(s(x), y) → f(f(x, y), 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(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)

Tuples:

F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

S tuples:

F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), 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(0, z0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)

Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

S tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) 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(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

We considered the (Usable) Rules:

f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)

And the Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

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

POL(0) = 0   
POL(F(x₁, x₂)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)

Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

S tuples:none
K tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty

(10) BOUNDS(1, 1)