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

0 CpxTRS
↳1 RcToIrcProof (BOTH BOUNDS(ID, ID), 6 ms)
↳2 CpxTRS
↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 138 ms)
↳6 CdtProblem
↳7 CdtInstantiationProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtUsableRulesProof (⇔, 0 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
↳14 CdtProblem
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 BOUNDS(1, 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, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(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(s([]))))

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

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

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

(5) 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(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), s(z1)) → c2(F(z0, s(c(s(z1)))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = [3]x2   
POL(c(x1)) = [3] + x1   
POL(c1(x1, x2)) = x1 + x2   
POL(c2(x1)) = x1   
POL(f(x1, x2)) = [3]x2   
POL(s(x1)) = [2]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

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

F(s(z0), s(z1)) → c2(F(z0, s(c(s(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), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

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

F(s(z0), s(z1)) → c2(F(z0, s(c(s(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), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

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

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(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), s(z1)) → c2(F(z0, s(c(s(z1)))))
We considered the (Usable) Rules:none
And the Tuples:

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