The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, we have rc = irc.

(2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))

Tuples:

REV(nil) → c
REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
CAR(.(z0, z1)) → c2
CDR(.(z0, z1)) → c3
NULL(nil) → c4
NULL(.(z0, z1)) → c5
++'(nil, z0) → c6
++'(.(z0, z1), z2) → c7(++'(z1, z2))

S tuples:

REV(nil) → c
REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
CAR(.(z0, z1)) → c2
CDR(.(z0, z1)) → c3
NULL(nil) → c4
NULL(.(z0, z1)) → c5
++'(nil, z0) → c6
++'(.(z0, z1), z2) → c7(++'(z1, z2))

K tuples:none
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, CAR, CDR, NULL, ++'

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7

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

Removed 6 trailing nodes:

REV(nil) → c
NULL(nil) → c4
++'(nil, z0) → c6
CDR(.(z0, z1)) → c3
CAR(.(z0, z1)) → c2
NULL(.(z0, z1)) → c5

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil))
car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false
++(nil, z0) → z0
++(.(z0, z1), z2) → .(z0, ++(z1, z2))

Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

S tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

K tuples:none
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

car(.(z0, z1)) → z0
cdr(.(z0, z1)) → z1
null(nil) → true
null(.(z0, z1)) → false

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

S tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

K tuples:none
Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

(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.

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))

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

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

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

POL(++(x₁, x₂)) = 0   
POL(++'(x₁, x₂)) = 0   
POL(.(x₁, x₂)) = [1] + x₂   
POL(REV(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(nil) = 0   
POL(rev(x₁)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

S tuples:

++'(.(z0, z1), z2) → c7(++'(z1, z2))

K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))

Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

++'(.(z0, z1), z2) → c7(++'(z1, z2))

We considered the (Usable) Rules:

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

And the Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

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

POL(++(x₁, x₂)) = x₁ + [2]x₂   
POL(++'(x₁, x₂)) = [1] + [2]x₁   
POL(.(x₁, x₂)) = [2] + x₂   
POL(REV(x₁)) = x₁²   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁)) = x₁   
POL(nil) = 0   
POL(rev(x₁)) = [2]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

S tuples:none
K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
++'(.(z0, z1), z2) → c7(++'(z1, z2))

Defined Rule Symbols:

rev, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

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

The set S is empty

(14) BOUNDS(1, 1)