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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 13 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 67 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 1)

(0) 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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1]
transitions:
s0(0) → 0
00() → 0
p0(0, 0, 0) → 1
p1(0, 0, 0) → 1
01() → 2
p1(2, 0, 0) → 1
p1(2, 0, 2) → 1
0 → 1
2 → 1

(2) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(z0, z1, s(z2)) → p(z0, z2, z1)
p(z0, s(z1), 0) → p(0, z1, z0)
p(z0, 0, 0) → z0
Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
P(z0, 0, 0) → c2
S tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
P(z0, 0, 0) → c2
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

P(z0, 0, 0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(z0, z1, s(z2)) → p(z0, z2, z1)
p(z0, s(z1), 0) → p(0, z1, z0)
p(z0, 0, 0) → z0
Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
S tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

P

Compound Symbols:

c, c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

p(z0, z1, s(z2)) → p(z0, z2, z1)
p(z0, s(z1), 0) → p(0, z1, z0)
p(z0, 0, 0) → z0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
S tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

P

Compound Symbols:

c, 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.

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
We considered the (Usable) Rules:none
And the Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(P(x1, x2, x3)) = x1 + x2 + x3   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(s(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
S tuples:none
K tuples:

P(z0, z1, s(z2)) → c(P(z0, z2, z1))
P(z0, s(z1), 0) → c1(P(0, z1, z0))
Defined Rule Symbols:none

Defined Pair Symbols:

P

Compound Symbols:

c, c1

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

The set S is empty

(12) BOUNDS(1, 1)