### (0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
reverse(nil) → nil
shuffle(nil) → nil
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
app(nil, z0) → z0
reverse(nil) → nil
shuffle(nil) → nil
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
less_leaves(z0, leaf) → false
less_leaves(leaf, cons(z0, z1)) → true
less_leaves(cons(z0, z1), cons(z2, z3)) → less_leaves(concat(z0, z1), concat(z2, z3))
Tuples:

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(nil, z0) → c4
APP(add(z0, z1), z2) → c5(APP(z1, z2))
REVERSE(nil) → c6
SHUFFLE(nil) → c8
CONCAT(leaf, z0) → c10
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(z0, leaf) → c12
LESS_LEAVES(leaf, cons(z0, z1)) → c13
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(nil, z0) → c4
APP(add(z0, z1), z2) → c5(APP(z1, z2))
REVERSE(nil) → c6
SHUFFLE(nil) → c8
CONCAT(leaf, z0) → c10
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(z0, leaf) → c12
LESS_LEAVES(leaf, cons(z0, z1)) → c13
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
K tuples:none
Defined Rule Symbols:

minus, quot, app, reverse, shuffle, concat, less_leaves

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14

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

Removed 8 trailing nodes:

CONCAT(leaf, z0) → c10
MINUS(z0, 0) → c
QUOT(0, s(z0)) → c2
APP(nil, z0) → c4
LESS_LEAVES(z0, leaf) → c12
REVERSE(nil) → c6
LESS_LEAVES(leaf, cons(z0, z1)) → c13
SHUFFLE(nil) → c8

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
app(nil, z0) → z0
reverse(nil) → nil
shuffle(nil) → nil
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
less_leaves(z0, leaf) → false
less_leaves(leaf, cons(z0, z1)) → true
less_leaves(cons(z0, z1), cons(z2, z3)) → less_leaves(concat(z0, z1), concat(z2, z3))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
K tuples:none
Defined Rule Symbols:

minus, quot, app, reverse, shuffle, concat, less_leaves

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
shuffle(nil) → nil
less_leaves(z0, leaf) → false
less_leaves(leaf, cons(z0, z1)) → true
less_leaves(cons(z0, z1), cons(z2, z3)) → less_leaves(concat(z0, z1), concat(z2, z3))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
K tuples:none
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (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.

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
We considered the (Usable) Rules:

concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
concat(leaf, z0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = 0
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = x1
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(REVERSE(x1)) = 0
POL(SHUFFLE(x1)) = 0
POL(app(x1, x2)) = [2]
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = [1] + x1 + x2
POL(leaf) = 0
POL(minus(x1, x2)) = [5] + [3]x2
POL(nil) = [2]
POL(reverse(x1)) = 0
POL(s(x1)) = [3]

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (9) 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.

QUOT(s(z0), s(z1)) → c3(QUOT(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:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = [2]x2
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = 0
POL(MINUS(x1, x2)) = [1]
POL(QUOT(x1, x2)) = [2]x1
POL(REVERSE(x1)) = 0
POL(SHUFFLE(x1)) = 0
POL(app(x1, x2)) = 0
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = [4] + [5]x2
POL(cons(x1, x2)) = 0
POL(leaf) = 0
POL(minus(x1, x2)) = x1
POL(nil) = [3]
POL(reverse(x1)) = [2]
POL(s(x1)) = [1] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (11) 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.

We considered the (Usable) Rules:

reverse(nil) → nil
app(nil, z0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = 0
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = 0
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(REVERSE(x1)) = 0
POL(SHUFFLE(x1)) = [2]x1
POL(add(x1, x2)) = [2] + x2
POL(app(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = [4]x2
POL(cons(x1, x2)) = 0
POL(leaf) = 0
POL(minus(x1, x2)) = [3] + [4]x2
POL(nil) = 0
POL(reverse(x1)) = x1
POL(s(x1)) = [1]

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (13) 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)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = 0
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = 0
POL(MINUS(x1, x2)) = x1
POL(QUOT(x1, x2)) = [2]x12
POL(REVERSE(x1)) = 0
POL(SHUFFLE(x1)) = 0
POL(app(x1, x2)) = 0
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = 0
POL(cons(x1, x2)) = 0
POL(leaf) = 0
POL(minus(x1, x2)) = [1] + x1
POL(nil) = 0
POL(reverse(x1)) = 0
POL(s(x1)) = [2] + x1

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (15) 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.

CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
We considered the (Usable) Rules:

concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
concat(leaf, z0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = 0
POL(CONCAT(x1, x2)) = [1] + x1 + x2
POL(LESS_LEAVES(x1, x2)) = x2 + [2]x22 + x12
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(REVERSE(x1)) = 0
POL(SHUFFLE(x1)) = 0
POL(app(x1, x2)) = 0
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = [1] + x1 + x2
POL(leaf) = 0
POL(minus(x1, x2)) = 0
POL(nil) = 0
POL(reverse(x1)) = 0
POL(s(x1)) = 0

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

APP(add(z0, z1), z2) → c5(APP(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (17) 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.

We considered the (Usable) Rules:

reverse(nil) → nil
app(nil, z0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = 0
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = 0
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(REVERSE(x1)) = [1] + [2]x1
POL(SHUFFLE(x1)) = x12
POL(add(x1, x2)) = [1] + x2
POL(app(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = 0
POL(cons(x1, x2)) = 0
POL(leaf) = 0
POL(minus(x1, x2)) = 0
POL(nil) = 0
POL(reverse(x1)) = x1
POL(s(x1)) = 0

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:

APP(add(z0, z1), z2) → c5(APP(z1, z2))
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

### (19) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

APP(add(z0, z1), z2) → c5(APP(z1, z2))
We considered the (Usable) Rules:

reverse(nil) → nil
app(nil, z0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APP(x1, x2)) = x1
POL(CONCAT(x1, x2)) = 0
POL(LESS_LEAVES(x1, x2)) = 0
POL(MINUS(x1, x2)) = 0
POL(QUOT(x1, x2)) = 0
POL(REVERSE(x1)) = [1] + x12
POL(SHUFFLE(x1)) = x12 + x13
POL(add(x1, x2)) = [1] + x1 + x2
POL(app(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c11(x1)) = x1
POL(c14(x1, x2, x3)) = x1 + x2 + x3
POL(c3(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(c9(x1, x2)) = x1 + x2
POL(concat(x1, x2)) = 0
POL(cons(x1, x2)) = 0
POL(leaf) = 0
POL(minus(x1, x2)) = 0
POL(nil) = 0
POL(reverse(x1)) = x1
POL(s(x1)) = 0

### (20) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
app(nil, z0) → z0
concat(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
S tuples:none
K tuples:

LESS_LEAVES(cons(z0, z1), cons(z2, z3)) → c14(LESS_LEAVES(concat(z0, z1), concat(z2, z3)), CONCAT(z0, z1), CONCAT(z2, z3))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
APP(add(z0, z1), z2) → c5(APP(z1, z2))
Defined Rule Symbols:

minus, reverse, app, concat

Defined Pair Symbols:

MINUS, QUOT, APP, REVERSE, SHUFFLE, CONCAT, LESS_LEAVES

Compound Symbols:

c1, c3, c5, c7, c9, c11, c14

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

The set S is empty