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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 14 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtUsableRulesProof (⇔, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 89 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 1 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳12 CdtProblem
        ↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 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(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

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:
f0(0, 0) → 0
h0(0, 0) → 0
g0(0, 0) → 1
g1(0, 0) → 2
f1(0, 2) → 1
f1(0, 0) → 3
g1(0, 3) → 1
g1(0, 0) → 4
h1(4, 0) → 1
g1(0, 3) → 2
f1(0, 2) → 2
f1(0, 2) → 4
f1(0, 3) → 3
g1(0, 3) → 4
h1(4, 0) → 2
h1(4, 0) → 4

(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:

g(f(z0, z1), z2) → f(z0, g(z1, z2))
g(h(z0, z1), z2) → g(z0, f(z1, z2))
g(z0, h(z1, z2)) → h(g(z0, z1), z2)
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(f(z0, z1), z2) → f(z0, g(z1, z2))
g(h(z0, z1), z2) → g(z0, f(z1, z2))
g(z0, h(z1, z2)) → h(g(z0, z1), z2)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(7) 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(z0, h(z1, z2)) → c2(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1, x2)) = [3]x2   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(f(x1, x2)) = 0   
POL(h(x1, x2)) = [3] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
K tuples:

G(z0, h(z1, z2)) → c2(G(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(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.

G(f(z0, z1), z2) → c(G(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1, x2)) = [2]x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(f(x1, x2)) = [2] + x2   
POL(h(x1, x2)) = x1 + x2   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:

G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
K tuples:

G(z0, h(z1, z2)) → c2(G(z0, z1))
G(f(z0, z1), z2) → c(G(z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

(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.

G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1, x2)) = x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(f(x1, x2)) = x2   
POL(h(x1, x2)) = [1] + x1 + x2   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
G(z0, h(z1, z2)) → c2(G(z0, z1))
S tuples:none
K tuples:

G(z0, h(z1, z2)) → c2(G(z0, z1))
G(f(z0, z1), z2) → c(G(z1, z2))
G(h(z0, z1), z2) → c1(G(z0, f(z1, z2)))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c, c1, c2

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

The set S is empty

(14) BOUNDS(1, 1)