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:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
f(s([]))
g(s([]), y)

The defined contexts are:
g(x0, [])
g(x0, s([]))
+([], s(x1))
+([], x1)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, 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:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

F(0) → c
F(s(z0)) → c1(G(z0, s(z0)))
G(0, z0) → c2
G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

F(0) → c
F(s(z0)) → c1(G(z0, s(z0)))
G(0, z0) → c2
G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

f, g, +

Defined Pair Symbols:

F, G, +'

Compound Symbols:

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

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

F(s(z0)) → c1(G(z0, s(z0)))

Removed 3 trailing nodes:

+'(z0, 0) → c5
F(0) → c
G(0, z0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

f, g, +

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → 1
f(s(z0)) → g(z0, s(z0))
g(0, z0) → z0
g(s(z0), z1) → g(z0, +(z1, s(z0)))
g(s(z0), z1) → g(z0, s(+(z1, z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, s(z1)) → s(+(z0, z1))
+(z0, 0) → z0

Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

(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.

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

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

POL(+(x₁, x₂)) = 0   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(G(x₁, x₂)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, s(z1)) → s(+(z0, z1))
+(z0, 0) → z0

Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))

Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

(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, s(z1)) → c6(+'(z0, z1))

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

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

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

POL(+(x₁, x₂)) = [2]x₁ + [2]x₂² + [2]x₁·x₂   
POL(+'(x₁, x₂)) = [2]x₂   
POL(0) = [2]   
POL(G(x₁, x₂)) = [2]x₁ + x₁²   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, s(z1)) → s(+(z0, z1))
+(z0, 0) → z0

Tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:none
K tuples:

G(s(z0), z1) → c3(G(z0, +(z1, s(z0))), +'(z1, s(z0)))
G(s(z0), z1) → c4(G(z0, s(+(z1, z0))), +'(z1, z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

Defined Rule Symbols:

+

Defined Pair Symbols:

G, +'

Compound Symbols:

c3, c4, c6

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

The set S is empty

(14) BOUNDS(1, 1)