### (0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^3).

The TRS R consists of the following rules:

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))

Rewrite Strategy: FULL

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

The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
*(x, *(y, z)) → *(otimes(x, y), z)

### (2) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^3).

The TRS R consists of the following rules:

*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

### (4) 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:

*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
*(1, z0) → z0
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2))
Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(1, z0) → c1
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
S tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(1, z0) → c1
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

*'(1, z0) → c1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
*(1, z0) → z0
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2))
Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
S tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c, c2

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
*(1, z0) → z0
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
S tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

*'

Compound Symbols:

c, c2

### (11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(*'(x1, x2)) = x1 + [2]x1·x2
POL(+(x1, x2)) = [1] + x1 + x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1, x2)) = x1 + x2
POL(oplus(x1, x2)) = [1] + x1 + x2

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
S tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
K tuples:

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

*'

Compound Symbols:

c, c2

### (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.

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(*'(x1, x2)) = x2 + x1·x2 + x1·x22
POL(+(x1, x2)) = [1] + x1 + x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1, x2)) = x1 + x2
POL(oplus(x1, x2)) = [1] + x1 + x2

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
S tuples:none
K tuples:

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2))
*'(z0, oplus(z1, z2)) → c(*'(z0, z1), *'(z0, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

*'

Compound Symbols:

c, c2

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

The set S is empty