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:

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

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:
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(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:

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

Rewrite Strategy: INNERMOST

(3) 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)

(4) 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))
rev(nil) → nil
++(nil, y) → y
++(x, nil) → x
make(x) → .(x, nil)

Rewrite Strategy: INNERMOST

(5) 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:
.0(0, 0) → 0
nil0() → 0
++0(0, 0) → 1
rev0(0) → 2
make0(0) → 3
++1(0, 0) → 4
.1(0, 4) → 1
nil1() → 2
nil1() → 5
.1(0, 5) → 3
.1(0, 4) → 4
0 → 1
0 → 4

(6) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))
++'(nil, z0) → c1
++'(z0, nil) → c2
REV(nil) → c3
MAKE(z0) → c4

S tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))
++'(nil, z0) → c1
++'(z0, nil) → c2
REV(nil) → c3
MAKE(z0) → c4

K tuples:none
Defined Rule Symbols:

++, rev, make

Defined Pair Symbols:

++', REV, MAKE

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 4 trailing nodes:

MAKE(z0) → c4
++'(z0, nil) → c2
++'(nil, z0) → c1
REV(nil) → c3

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

S tuples:

++'(.(z0, z1), z2) → c(++'(z1, z2))

K tuples:none
Defined Rule Symbols:

++, rev, make

Defined Pair Symbols:

++'

Compound Symbols:

c

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(12) 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

(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.

++'(.(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   

(14) 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

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

The set S is empty

(16) BOUNDS(1, 1)