(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:

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
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
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: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
quot(s(x), s([]))

The defined contexts are:
quot([], s(x1))
shuffle([])
less_leaves([], x1)
less_leaves(x0, [])
app([], add(x1, nil))
minus([], x1)
app([], x1)
reverse([])
concat([], x1)
concat(x0, [])
app(x0, add([], nil))
app(x0, [])

[] just represents basic- or constructor-terms in the following defined contexts:
quot([], 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^3).


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
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
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

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

Converted Cpx (relative) TRS to CDT

(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
app(add(z0, z1), z2) → add(z0, app(z1, z2))
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
shuffle(nil) → nil
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))
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
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(nil) → c8
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(nil) → c8
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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

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

Removed 8 trailing nodes:

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

(6) 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
app(add(z0, z1), z2) → add(z0, app(z1, z2))
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
shuffle(nil) → nil
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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

(7) 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
shuffle(add(z0, z1)) → add(z0, shuffle(reverse(z1)))
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))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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

(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.

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(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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)) = [2]x1   
POL(MINUS(x1, x2)) = 0   
POL(QUOT(x1, x2)) = 0   
POL(REVERSE(x1)) = 0   
POL(SHUFFLE(x1)) = 0   
POL(add(x1, x2)) = 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   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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

(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.

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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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)) = x1   
POL(REVERSE(x1)) = 0   
POL(SHUFFLE(x1)) = 0   
POL(add(x1, x2)) = 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)) = x1   
POL(nil) = 0   
POL(reverse(x1)) = 0   
POL(s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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(n^1)) transformation)

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

SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
We considered the (Usable) Rules:

reverse(nil) → nil
app(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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)) = x1   
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   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
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))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(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(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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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) = [2]   
POL(APP(x1, x2)) = x2 + x22   
POL(CONCAT(x1, x2)) = 0   
POL(LESS_LEAVES(x1, x2)) = 0   
POL(MINUS(x1, x2)) = [2] + x1   
POL(QUOT(x1, x2)) = x1·x2 + x12   
POL(REVERSE(x1)) = 0   
POL(SHUFFLE(x1)) = 0   
POL(add(x1, x2)) = 0   
POL(app(x1, x2)) = x1 + x2 + x22 + [2]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)) = [2] + [2]x1 + [2]x22 + x1·x2   
POL(cons(x1, x2)) = 0   
POL(leaf) = [2]   
POL(minus(x1, x2)) = x1   
POL(nil) = 0   
POL(reverse(x1)) = 0   
POL(s(x1)) = [1] + x1   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
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))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(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

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(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(leaf, z0) → z0
concat(cons(z0, z1), z2) → cons(z0, concat(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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) = [1]   
POL(APP(x1, x2)) = x2   
POL(CONCAT(x1, x2)) = [1] + x1   
POL(LESS_LEAVES(x1, x2)) = x2 + x22 + [2]x12   
POL(MINUS(x1, x2)) = 0   
POL(QUOT(x1, x2)) = 0   
POL(REVERSE(x1)) = 0   
POL(SHUFFLE(x1)) = 0   
POL(add(x1, x2)) = 0   
POL(app(x1, x2)) = [1] + x1 + [2]x2 + [2]x22 + [2]x1·x2 + [2]x12   
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)) = [1] + x1 + x2   
POL(cons(x1, x2)) = [2] + x1 + x2   
POL(leaf) = [1]   
POL(minus(x1, x2)) = [2] + x2 + x22   
POL(nil) = 0   
POL(reverse(x1)) = 0   
POL(s(x1)) = [1]   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
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))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(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(n^2)) transformation)

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

REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
We considered the (Usable) Rules:

reverse(nil) → nil
app(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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) = [2]   
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)) = [2] + [2]x22 + x1·x2 + x12   
POL(cons(x1, x2)) = 0   
POL(leaf) = [2]   
POL(minus(x1, x2)) = [2] + [2]x2 + [2]x22 + x12   
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
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(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

(21) CdtRuleRemovalProof (UPPER BOUND(ADD(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(add(z0, z1), z2) → add(z0, app(z1, z2))
app(nil, z0) → z0
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
reverse(nil) → nil
reverse(add(z0, z1)) → app(reverse(z1), add(z0, nil))
app(nil, z0) → z0
app(add(z0, z1), z2) → add(z0, app(z1, z2))
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))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
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))
SHUFFLE(add(z0, z1)) → c9(SHUFFLE(reverse(z1)), REVERSE(z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
CONCAT(cons(z0, z1), z2) → c11(CONCAT(z1, z2))
REVERSE(add(z0, z1)) → c7(APP(reverse(z1), add(z0, nil)), REVERSE(z1))
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

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

The set S is empty

(24) BOUNDS(1, 1)