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

(0) 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))

Rewrite Strategy: INNERMOST

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

The following defined symbols can occur below the 0th argument of quot: minus
The following defined symbols can occur below the 0th argument of minus: minus

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))

(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:

quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(s(x), y) → s(plus(x, y))
plus(0, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
quot(0, s(y)) → 0

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(0, z0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c1
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
PLUS(0, z0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MINUS(z0, 0) → c5

S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
QUOT(0, s(z0)) → c1
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
PLUS(0, z0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MINUS(z0, 0) → c5

K tuples:none
Defined Rule Symbols:

quot, plus, minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

PLUS(0, z0) → c3
MINUS(z0, 0) → c5
QUOT(0, s(z0)) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(0, z0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

quot, plus, minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
quot(0, s(z0)) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(0, z0) → z0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

(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.

PLUS(s(z0), z1) → c2(PLUS(z0, z1))

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

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

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

POL(0) = 0   
POL(MINUS(x₁, x₂)) = 0   
POL(PLUS(x₁, x₂)) = x₁   
POL(QUOT(x₁, x₂)) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(minus(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

S tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))

Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

(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.

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

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

POL(0) = 0   
POL(MINUS(x₁, x₂)) = 0   
POL(PLUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

S tuples:

MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

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

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

MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

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

POL(0) = [2]   
POL(MINUS(x₁, x₂)) = [2] + x₂   
POL(PLUS(x₁, x₂)) = [2]x₁   
POL(QUOT(x₁, x₂)) = [2]x₁·x₂   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [2] + x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
PLUS(s(z0), z1) → c2(PLUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

S tuples:none
K tuples:

PLUS(s(z0), z1) → c2(PLUS(z0, z1))
QUOT(s(z0), s(z1)) → c(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

Defined Rule Symbols:

minus

Defined Pair Symbols:

QUOT, PLUS, MINUS

Compound Symbols:

c, c2, c4

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

The set S is empty

(16) BOUNDS(1, 1)