### (0) Obligation:

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

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: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
mod(s(x), s([]))
mod(s([]), s(y))
if_mod(true, s(x), s([]))

The defined contexts are:
if_mod([], s(x1), s(x2))
mod([], s(x1))
le(x0, [])
if_mod(x0, s([]), s(x2))
minus([], x1)

[] just represents basic- or constructor-terms in the following defined contexts:
if_mod([], s(x1), s(x2))
mod([], s(x1))

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (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:

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

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

Converted Cpx (relative) TRS to CDT

### (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(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

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

Removed 6 trailing nodes:

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

### (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)
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

### (7) 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)

### (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))
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

### (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.

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) = 0
POL(IF_MOD(x1, x2, x3)) = x2
POL(LE(x1, x2)) = 0
POL(MINUS(x1, x2)) = 0
POL(MOD(x1, x2)) = x1
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

### (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))
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))
K tuples:

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

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

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

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))

### (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:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:

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

le, minus

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

### (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.

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

MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:

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

### (15) 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:

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)) = x22
POL(LE(x1, x2)) = 0
POL(MINUS(x1, x2)) = [1] + x1
POL(MOD(x1, x2)) = 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

### (16) 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:

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

### (17) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty