The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^3).

(0) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^3).


The TRS R consists of the following rules:

app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
shuffle(nil) → nil
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))

Tuples:

APP(nil, z0) → c
APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(nil) → c2
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(nil) → c4
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:

APP(nil, z0) → c
APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(nil) → c2
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(nil) → c4
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

K tuples:none
Defined Rule Symbols:

app, reverse, shuffle

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

SHUFFLE(nil) → c4
REVERSE(nil) → c2
APP(nil, z0) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
shuffle(nil) → nil
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))

Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

K tuples:none
Defined Rule Symbols:

app, reverse, shuffle

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c1, c3, c5

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

shuffle(nil) → nil
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))

Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

K tuples:none
Defined Rule Symbols:

reverse, app

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c1, c3, c5

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

We considered the (Usable) Rules:

reverse(nil) → nil
app(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))

And the Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

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

POL(APP(x₁, x₂)) = 0   
POL(REVERSE(x₁)) = 0   
POL(SHUFFLE(x₁)) = x₁   
POL(add(x₁, x₂)) = [2] + x₂   
POL(app(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   
POL(reverse(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))

Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))

K tuples:

SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

Defined Rule Symbols:

reverse, app

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c1, c3, c5

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

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

REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))

We considered the (Usable) Rules:

reverse(nil) → nil
app(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))

And the Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

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

POL(APP(x₁, x₂)) = 0   
POL(REVERSE(x₁)) = [1] + x₁   
POL(SHUFFLE(x₁)) = [2]x₁²   
POL(add(x₁, x₂)) = [1] + x₁ + x₂   
POL(app(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   
POL(reverse(x₁)) = x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))

Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))

K tuples:

SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))

Defined Rule Symbols:

reverse, app

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c1, c3, c5

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)) transformation)

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

APP(add(z0, z1), z2) → c1(APP(z1, z2))

We considered the (Usable) Rules:

reverse(nil) → nil
app(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))

And the Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

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

POL(APP(x₁, x₂)) = x₁ + x₁·x₂   
POL(REVERSE(x₁)) = [1] + x₁²   
POL(SHUFFLE(x₁)) = x₁ + x₁² + x₁³   
POL(add(x₁, x₂)) = [1] + x₂   
POL(app(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   
POL(reverse(x₁)) = x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))

Tuples:

APP(add(z0, z1), z2) → c1(APP(z1, z2))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))

S tuples:none
K tuples:

SHUFFLE(add(z0, z1)) → c5(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(add(z0, z1)) → c3(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
APP(add(z0, z1), z2) → c1(APP(z1, z2))

Defined Rule Symbols:

reverse, app

Defined Pair Symbols:

APP, REVERSE, SHUFFLE

Compound Symbols:

c1, c3, c5

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

The set S is empty

(14) BOUNDS(1, 1)