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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))

Tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(0, z0) → c3
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(true, s(z0), z1) → c5
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(0, s(z0)) → c7
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(0, z0) → c3
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(true, s(z0), z1) → c5
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(0, s(z0)) → c7
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8

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

Removed 5 trailing nodes:

MINUS(0, z0) → c3
QUOT(0, s(z0)) → c7
LE(s(z0), 0) → c1
LE(0, z0) → c
IFMINUS(true, s(z0), z1) → c5

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true
le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

le, minus, ifMinus, quot

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

le, minus, ifMinus

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

(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.

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

We considered the (Usable) Rules:

ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0

And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = 0   
POL(IFMINUS(x₁, x₂, x₃)) = 0   
POL(LE(x₁, x₂)) = 0   
POL(MINUS(x₁, x₂)) = 0   
POL(QUOT(x₁, x₂)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(ifMinus(x₁, x₂, x₃)) = x₂   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(true) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))

K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

Defined Rule Symbols:

le, minus, ifMinus

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

(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.

IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))

We considered the (Usable) Rules:

ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0

And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = 0   
POL(IFMINUS(x₁, x₂, x₃)) = x₂   
POL(LE(x₁, x₂)) = 0   
POL(MINUS(x₁, x₂)) = x₁   
POL(QUOT(x₁, x₂)) = x₁²   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(ifMinus(x₁, x₂, x₃)) = x₂   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(true) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))

K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))

Defined Rule Symbols:

le, minus, ifMinus

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))

K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))

Defined Rule Symbols:

le, minus, ifMinus

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

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

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

LE(s(z0), s(z1)) → c2(LE(z0, z1))

We considered the (Usable) Rules:

ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
minus(0, z0) → 0

And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = 0   
POL(IFMINUS(x₁, x₂, x₃)) = [1] + x₂²   
POL(LE(x₁, x₂)) = x₁   
POL(MINUS(x₁, x₂)) = [1] + x₁ + x₁²   
POL(QUOT(x₁, x₂)) = x₁³   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(ifMinus(x₁, x₂, x₃)) = x₂   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(true) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(s(z0), 0) → false
le(s(z0), s(z1)) → le(z0, z1)
le(0, z0) → true
minus(0, z0) → 0
minus(s(z0), z1) → ifMinus(le(s(z0), z1), s(z0), z1)
ifMinus(true, s(z0), z1) → 0
ifMinus(false, s(z0), z1) → s(minus(z0, z1))

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:none
K tuples:

QUOT(s(z0), s(z1)) → c8(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
IFMINUS(false, s(z0), z1) → c6(MINUS(z0, z1))
MINUS(s(z0), z1) → c4(IFMINUS(le(s(z0), z1), s(z0), z1), LE(s(z0), z1))
LE(s(z0), s(z1)) → c2(LE(z0, z1))

Defined Rule Symbols:

le, minus, ifMinus

Defined Pair Symbols:

LE, MINUS, IFMINUS, QUOT

Compound Symbols:

c2, c4, c6, c8

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

The set S is empty

(16) BOUNDS(1, 1)