Runtime Complexity (innermost) proof of /tmp/tmpYSxft4/#4.26.xml

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

0 CpxTRS

↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)

↳2 CdtProblem

↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)

↳4 CdtProblem

↳5 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 6 ms)

↳6 CdtProblem

↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)

↳8 CdtProblem

↳9 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)

↳10 CdtProblem

↳11 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)

↳12 CdtProblem

↳13 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)

↳14 CdtProblem

↳15 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 13 ms)

↳16 CdtProblem

↳17 CdtKnowledgeProof (⇔, 0 ms)

↳18 CdtProblem

↳19 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 122 ms)

↳20 CdtProblem

↳21 SIsEmptyProof (⇔, 0 ms)

↳22 BOUNDS(O(1), O(1))

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, z1) → c5(IF(le(z0, z1), z0, z1), LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1), P(z0))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, z1) → c5(IF(le(z0, z1), z0, z1), LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1), P(z0))

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c7

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, z1) → c5(IF(le(z0, z1), z0, z1), LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(z0, z1) → c5(IF(le(z0, z1), z0, z1), LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c7

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

Use narrowing to replace MINUS(z0, z1) → c5(IF(le(z0, z1), z0, z1), LE(z0, z1)) by

MINUS(0, z0) → c5(IF(true, 0, z0), LE(0, z0))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0), LE(s(z0), 0))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(x0, x1) → c5

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(0, z0) → c5(IF(true, 0, z0), LE(0, z0))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0), LE(s(z0), 0))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(x0, x1) → c5

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(0, z0) → c5(IF(true, 0, z0), LE(0, z0))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0), LE(s(z0), 0))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(x0, x1) → c5

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, IF, MINUS

Compound Symbols:

c4, c7, c5, c5

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

Removed 2 trailing nodes:

MINUS(0, z0) → c5(IF(true, 0, z0), LE(0, z0))
MINUS(x0, x1) → c5

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0), LE(s(z0), 0))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0), LE(s(z0), 0))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, IF, MINUS

Compound Symbols:

c4, c7, c5

(9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
IF(false, z0, z1) → c7(MINUS(p(z0), z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, IF, MINUS

Compound Symbols:

c4, c7, c5, c5

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

Use narrowing to replace IF(false, z0, z1) → c7(MINUS(p(z0), z1)) by

IF(false, 0, x1) → c7(MINUS(0, x1))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))
IF(false, x0, x1) → c7

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, 0, x1) → c7(MINUS(0, x1))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))
IF(false, x0, x1) → c7

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, 0, x1) → c7(MINUS(0, x1))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))
IF(false, x0, x1) → c7

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c5, c7, c7

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

Removed 2 trailing nodes:

IF(false, 0, x1) → c7(MINUS(0, x1))
IF(false, x0, x1) → c7

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

K tuples:none
Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c5, c7

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

IF(false, s(z0), x1) → c7(MINUS(z0, x1))

We considered the (Usable) Rules:

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

And the Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

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

POL(0) = 0
POL(IF(x₁, x₂, x₃)) = x₂
POL(LE(x₁, x₂)) = 0
POL(MINUS(x₁, x₂)) = x₁
POL(c4(x₁)) = x₁
POL(c5(x₁)) = x₁
POL(c5(x₁, x₂)) = x₁ + x₂
POL(c7(x₁)) = x₁
POL(false) = 0
POL(le(x₁, x₂)) = [3] + [4]x₁ + [2]x₂
POL(s(x₁)) = [2] + x₁
POL(true) = [2]

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))

K tuples:

IF(false, s(z0), x1) → c7(MINUS(z0, x1))

Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c5, c7

(17) CdtKnowledgeProof (EQUIVALENT transformation)

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

MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

S tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))

K tuples:

IF(false, s(z0), x1) → c7(MINUS(z0, x1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))

Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c5, c7

(19) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

LE(s(z0), s(z1)) → c4(LE(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

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

POL(0) = 0
POL(IF(x₁, x₂, x₃)) = [2]x₂·x₃
POL(LE(x₁, x₂)) = x₂
POL(MINUS(x₁, x₂)) = [2]x₂ + [2]x₁·x₂
POL(c4(x₁)) = x₁
POL(c5(x₁)) = x₁
POL(c5(x₁, x₂)) = x₁ + x₂
POL(c7(x₁)) = x₁
POL(false) = 0
POL(le(x₁, x₂)) = 0
POL(s(x₁)) = [2] + x₁
POL(true) = 0

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c4(LE(z0, z1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
IF(false, s(z0), x1) → c7(MINUS(z0, x1))

S tuples:none
K tuples:

IF(false, s(z0), x1) → c7(MINUS(z0, x1))
MINUS(s(z0), s(z1)) → c5(IF(le(z0, z1), s(z0), s(z1)), LE(s(z0), s(z1)))
MINUS(s(z0), 0) → c5(IF(false, s(z0), 0))
LE(s(z0), s(z1)) → c4(LE(z0, z1))

Defined Rule Symbols:

p, le, minus, if

Defined Pair Symbols:

LE, MINUS, IF

Compound Symbols:

c4, c5, c5, c7

(21) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(0) Obligation:

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

(2) Obligation:

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

(4) Obligation:

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

(6) Obligation:

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

(8) Obligation:

(9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

(10) Obligation:

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

(12) Obligation:

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

(14) Obligation:

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

(16) Obligation:

(17) CdtKnowledgeProof (EQUIVALENT transformation)

(18) Obligation:

(19) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

(20) Obligation:

(21) SIsEmptyProof (EQUIVALENT transformation)

(22) BOUNDS(O(1), O(1))