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:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
mod(0, z0) → 0
mod(s(z0), 0) → 0
mod(s(z0), s(z1)) → if(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0))

Tuples:

LEQ(0, z0) → c
LEQ(s(z0), 0) → c1
LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
IF(true, z0, z1) → c3
IF(false, z0, z1) → c4
-'(z0, 0) → c5
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(0, z0) → c7
MOD(s(z0), 0) → c8
MOD(s(z0), s(z1)) → c9(IF(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0)), LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(0, z0) → c
LEQ(s(z0), 0) → c1
LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
IF(true, z0, z1) → c3
IF(false, z0, z1) → c4
-'(z0, 0) → c5
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(0, z0) → c7
MOD(s(z0), 0) → c8
MOD(s(z0), s(z1)) → c9(IF(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0)), LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

leq, if, -, mod

Defined Pair Symbols:

LEQ, IF, -', MOD

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

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

Removed 7 trailing nodes:

MOD(s(z0), 0) → c8
LEQ(0, z0) → c
-'(z0, 0) → c5
MOD(0, z0) → c7
IF(false, z0, z1) → c4
LEQ(s(z0), 0) → c1
IF(true, z0, z1) → c3

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
mod(0, z0) → 0
mod(s(z0), 0) → 0
mod(s(z0), s(z1)) → if(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0))

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(IF(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0)), LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(IF(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0)), LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

leq, if, -, mod

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
mod(0, z0) → 0
mod(s(z0), 0) → 0
mod(s(z0), s(z1)) → if(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0))

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

leq, if, -, mod

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

leq(0, z0) → true
leq(s(z0), 0) → false
leq(s(z0), s(z1)) → leq(z0, z1)
if(true, z0, z1) → z0
if(false, z0, z1) → z1
mod(0, z0) → 0
mod(s(z0), 0) → 0
mod(s(z0), s(z1)) → if(leq(z1, z0), mod(-(s(z0), s(z1)), s(z1)), s(z0))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

-

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

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

Use narrowing to replace MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(s(z0), s(z1)), s(z1)), -'(s(z0), s(z1))) by

MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

K tuples:none
Defined Rule Symbols:

-

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

(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.

MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

We considered the (Usable) Rules:

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

And the Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

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

POL(-(x₁, x₂)) = x₁   
POL(-'(x₁, x₂)) = [1]   
POL(0) = 0   
POL(LEQ(x₁, x₂)) = 0   
POL(MOD(x₁, x₂)) = x₁   
POL(c2(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c9(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(s(x₁)) = [2] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

S tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

K tuples:

MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

Defined Rule Symbols:

-

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

(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.

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

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

POL(-(x₁, x₂)) = x₁   
POL(-'(x₁, x₂)) = [2]x₁   
POL(0) = 0   
POL(LEQ(x₁, x₂)) = [1] + [2]x₂   
POL(MOD(x₁, x₂)) = [2]x₁·x₂ + x₁²   
POL(c2(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c9(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(s(x₁)) = [2] + x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))
MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))

S tuples:none
K tuples:

MOD(s(z0), s(z1)) → c9(LEQ(z1, z0), MOD(-(z0, z1), s(z1)), -'(s(z0), s(z1)))
LEQ(s(z0), s(z1)) → c2(LEQ(z0, z1))
-'(s(z0), s(z1)) → c6(-'(z0, z1))

Defined Rule Symbols:

-

Defined Pair Symbols:

LEQ, -', MOD

Compound Symbols:

c2, c6, c9

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

The set S is empty

(16) BOUNDS(1, 1)