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

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(x, y), v))

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:
nil0() → 0
++0(0, 0) → 0
u0() → 0
v0() → 0
merge0(0, 0) → 1
u1() → 4
v1() → 5
++1(4, 5) → 3
merge1(0, 3) → 2
++1(0, 2) → 1
++1(0, 0) → 7
v1() → 8
merge1(7, 8) → 6
++1(4, 6) → 1
++1(0, 2) → 2
++1(4, 6) → 2
0 → 1
3 → 2

(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:

merge(z0, nil) → z0
merge(nil, z0) → z0
merge(++(z0, z1), ++(u, v)) → ++(z0, merge(z1, ++(u, v)))
merge(++(z0, z1), ++(u, v)) → ++(u, merge(++(z0, z1), v))

Tuples:

MERGE(z0, nil) → c
MERGE(nil, z0) → c1
MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))
MERGE(++(z0, z1), ++(u, v)) → c3(MERGE(++(z0, z1), v))

S tuples:

MERGE(z0, nil) → c
MERGE(nil, z0) → c1
MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))
MERGE(++(z0, z1), ++(u, v)) → c3(MERGE(++(z0, z1), v))

K tuples:none
Defined Rule Symbols:

merge

Defined Pair Symbols:

MERGE

Compound Symbols:

c, c1, c2, c3

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

Removed 3 trailing nodes:

MERGE(nil, z0) → c1
MERGE(++(z0, z1), ++(u, v)) → c3(MERGE(++(z0, z1), v))
MERGE(z0, nil) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

merge(z0, nil) → z0
merge(nil, z0) → z0
merge(++(z0, z1), ++(u, v)) → ++(z0, merge(z1, ++(u, v)))
merge(++(z0, z1), ++(u, v)) → ++(u, merge(++(z0, z1), v))

Tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

S tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

K tuples:none
Defined Rule Symbols:

merge

Defined Pair Symbols:

MERGE

Compound Symbols:

c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

merge(z0, nil) → z0
merge(nil, z0) → z0
merge(++(z0, z1), ++(u, v)) → ++(z0, merge(z1, ++(u, v)))
merge(++(z0, z1), ++(u, v)) → ++(u, merge(++(z0, z1), v))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

S tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MERGE

Compound Symbols:

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.

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

We considered the (Usable) Rules:none
And the Tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(++(x₁, x₂)) = [1] + x₂   
POL(MERGE(x₁, x₂)) = x₁   
POL(c2(x₁)) = x₁   
POL(u) = 0   
POL(v) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

S tuples:none
K tuples:

MERGE(++(z0, z1), ++(u, v)) → c2(MERGE(z1, ++(u, v)))

Defined Rule Symbols:none

Defined Pair Symbols:

MERGE

Compound Symbols:

c2

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

The set S is empty

(12) BOUNDS(1, 1)