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

0 CpxTRS
↳1 DependencyGraphProof (BOTH BOUNDS(ID, ID), 12 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 0 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳6 CdtProblem
        ↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtUsableRulesProof (⇔, 0 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 91 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 (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

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

Rewrite Strategy: INNERMOST

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

The following rules are not reachable from basic terms in the dependency graph and can be removed:
f(f(x)) → f(d(f(x)))

(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:

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

Rewrite Strategy: INNERMOST

(3) 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, 2]
transitions:
c0(0, 0) → 0
s0(0) → 0
g0(0) → 1
f0(0) → 2
s1(0) → 4
c1(4, 0) → 3
g1(3) → 1
s1(0) → 6
c1(0, 6) → 5
f1(5) → 2
s1(4) → 4
s1(6) → 6
0 → 2
5 → 2

(4) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(z0) → c2
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(z0) → c2
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c2, c3

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

Removed 1 trailing nodes:

F(z0) → c2

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c3

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c3

(11) 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(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
We considered the (Usable) Rules:none
And the Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = x1   
POL(G(x1)) = 0   
POL(c(x1, x2)) = x1   
POL(c1(x1)) = x1   
POL(c3(x1)) = x1   
POL(s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
S tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
K tuples:

F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
Defined Rule Symbols:none

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c3

(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.

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

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1)) = 0   
POL(G(x1)) = x1   
POL(c(x1, x2)) = x2   
POL(c1(x1)) = x1   
POL(c3(x1)) = x1   
POL(s(x1)) = [1] + x1   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
S tuples:none
K tuples:

F(c(s(z0), z1)) → c3(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c1(G(c(s(z0), z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c3

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

The set S is empty

(16) BOUNDS(1, 1)