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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 7 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)), 102 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 6 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 8 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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

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, 2, 3]
transitions:
s0(0) → 0
cons0(0, 0) → 0
00() → 0
f0(0) → 1
g0(0) → 2
h0(0) → 3
f1(0) → 1
g1(0) → 2
s1(0) → 2
cons1(0, 0) → 5
g1(5) → 4
h1(4) → 3
g1(0) → 4
s1(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:

f(s(z0)) → f(z0)
g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
h(cons(z0, z1)) → h(g(cons(z0, z1)))
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
G(cons(s(z0), z1)) → c2
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
G(cons(s(z0), z1)) → c2
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, c1, c2, c3

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

Removed 1 trailing nodes:

G(cons(s(z0), z1)) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(s(z0)) → f(z0)
g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
h(cons(z0, z1)) → h(g(cons(z0, z1)))
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
K tuples:none
Defined Rule Symbols:

f, g, h

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, c1, c3

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(s(z0)) → f(z0)
h(cons(z0, z1)) → h(g(cons(z0, z1)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, c1, c3

(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.

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

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:

G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
K tuples:

F(s(z0)) → c(F(z0))
Defined Rule Symbols:

g

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, 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.

H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
We considered the (Usable) Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
And the Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(F(x1)) = 0   
POL(G(x1)) = [1]   
POL(H(x1)) = x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = [2] + x1 + x2   
POL(g(x1)) = 0   
POL(s(x1)) = 0   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:

G(cons(0, z0)) → c1(G(z0))
K tuples:

F(s(z0)) → c(F(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
Defined Rule Symbols:

g

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, 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(cons(0, z0)) → c1(G(z0))
We considered the (Usable) Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
And the Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(cons(0, z0)) → g(z0)
g(cons(s(z0), z1)) → s(z0)
Tuples:

F(s(z0)) → c(F(z0))
G(cons(0, z0)) → c1(G(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
S tuples:none
K tuples:

F(s(z0)) → c(F(z0))
H(cons(z0, z1)) → c3(H(g(cons(z0, z1))), G(cons(z0, z1)))
G(cons(0, z0)) → c1(G(z0))
Defined Rule Symbols:

g

Defined Pair Symbols:

F, G, H

Compound Symbols:

c, c1, c3

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

The set S is empty

(16) BOUNDS(1, 1)