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

(0) Obligation:

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


The TRS R consists of the following rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(X)) → X

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, we have rc = irc.

(2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, 1).


The TRS R consists of the following rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(X)) → 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(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(z0)) → z0

Tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(z0)) → c2

S tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(z0)) → c2

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F, P

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

P(s(z0)) → c2
F(0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(z0)) → z0

Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))

S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(z0)) → z0

Tuples:

F(s(0)) → c1(F(p(s(0))))

S tuples:

F(s(0)) → c1(F(p(s(0))))

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0

Tuples:

F(s(0)) → c1(F(p(s(0))))

S tuples:

F(s(0)) → c1(F(p(s(0))))

K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Use narrowing to replace F(s(0)) → c1(F(p(s(0)))) by

F(s(0)) → c1(F(0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0

Tuples:

F(s(0)) → c1(F(0))

S tuples:

F(s(0)) → c1(F(0))

K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing nodes:

F(s(0)) → c1(F(0))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0

Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:none

Compound Symbols:none

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

The set S is empty

(16) BOUNDS(1, 1)