The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

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

The defined contexts are:
f(x0, s([]))
f(x0, s(c([])))

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^2).


The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(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, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), z1) → f(z0, s(c(z1)))

Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

S tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

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

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

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

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

POL(F(x₁, x₂)) = x₂   
POL(c(x₁)) = [1] + x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(f(x₁, x₂)) = x₂ + x₂² + x₁·x₂   
POL(s(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

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

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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

Use instantiation to replace F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) by

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

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

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))

K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(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.

F(s(z0), z1) → c2(F(z0, s(c(z1))))

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

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

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

POL(F(x₁, x₂)) = x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
F(c(z1), c(z1)) → c1(F(z1, z1))

S tuples:none
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
F(s(z0), z1) → c2(F(z0, s(c(z1))))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

The set S is empty

(16) BOUNDS(1, 1)