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

(0) Obligation:

Runtime Complexity TRS:
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(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), 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:

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

Tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(z0, 0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(0, z0) → c5
MOD(s(z0), 0) → c6
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
IF_MOD(false, s(z0), s(z1)) → c9

S tuples:

LE(0, z0) → c
LE(s(z0), 0) → c1
LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(z0, 0) → c3
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(0, z0) → c5
MOD(s(z0), 0) → c6
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
IF_MOD(false, s(z0), s(z1)) → c9

K tuples:none
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

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

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

Removed 6 trailing nodes:

LE(0, z0) → c
LE(s(z0), 0) → c1
MOD(0, z0) → c5
MINUS(z0, 0) → c3
IF_MOD(false, s(z0), s(z1)) → c9
MOD(s(z0), 0) → c6

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

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

le, minus

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(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)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(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:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = [2]   
POL(IF_MOD(x₁, x₂, x₃)) = [2]x₂ + [2]x₃   
POL(LE(x₁, x₂)) = 0   
POL(MINUS(x₁, x₂)) = [3]   
POL(MOD(x₁, x₂)) = [2] + [2]x₁ + [2]x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = [5]   
POL(le(x₁, x₂)) = [3]x₂   
POL(minus(x₁, x₂)) = [1] + x₁   
POL(s(x₁)) = [4] + x₁   
POL(true) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

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

K tuples:

MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

Defined Rule Symbols:

le, minus

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

(9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(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:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = 0   
POL(IF_MOD(x₁, x₂, x₃)) = [2]x₂·x₃   
POL(LE(x₁, x₂)) = 0   
POL(MINUS(x₁, x₂)) = [3] + x₂   
POL(MOD(x₁, x₂)) = [1] + [2]x₁·x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [2] + x₁   
POL(true) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))

K tuples:

MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))

Defined Rule Symbols:

le, minus

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

(11) CdtRuleRemovalProof (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)) → c2(LE(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = 0   
POL(IF_MOD(x₁, x₂, x₃)) = x₂ + [2]x₂²   
POL(LE(x₁, x₂)) = [2] + x₂   
POL(MINUS(x₁, x₂)) = [2] + [3]x₁   
POL(MOD(x₁, x₂)) = [1] + [2]x₁ + [2]x₁²   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(true) = 0   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:none
K tuples:

MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
LE(s(z0), s(z1)) → c2(LE(z0, z1))

Defined Rule Symbols:

le, minus

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

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

The set S is empty

(14) BOUNDS(O(1), O(1))