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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

Rewrite Strategy: INNERMOST

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

The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
++(++(x, y), z) → ++(x, ++(y, z))

(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:

++(.(x, y), z) → .(x, ++(y, z))
++(nil, y) → y
++(x, nil) → x

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]
transitions:
.0(0, 0) → 0
nil0() → 0
++0(0, 0) → 1
++1(0, 0) → 2
.1(0, 2) → 1
.1(0, 2) → 2
0 → 1
0 → 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:

++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(nil, z0) → z0
++(z0, nil) → z0

Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))
++'(nil, z0) → c1
++'(z0, nil) → c2

S tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))
++'(nil, z0) → c1
++'(z0, nil) → c2

K tuples:none
Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

++'(z0, nil) → c2
++'(nil, z0) → c1

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(nil, z0) → z0
++(z0, nil) → z0

Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

S tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

K tuples:none
Defined Rule Symbols:

++

Defined Pair Symbols:

++'

Compound Symbols:

c

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

++(.(z0, z1), z2) → .(z0, ++(z1, z2))
++(nil, z0) → z0
++(z0, nil) → z0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

S tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

++'

Compound Symbols:

c

(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.

++'(.(z0, z1), z2) → c(++'(z1, z2))

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

++'(.(z0, z1), z2) → c(++'(z1, z2))

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

POL(++'(x₁, x₂)) = x₁   
POL(.(x₁, x₂)) = [1] + x₂   
POL(c(x₁)) = x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

S tuples:none
K tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

Defined Rule Symbols:none

Defined Pair Symbols:

++'

Compound Symbols:

c

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

The set S is empty

(14) BOUNDS(1, 1)