Editing Associative property (section)

==Generalized associative law==
[[Image:Tamari lattice.svg|thumb|In the absence of the associative property, five factors {{mvar|a}}, {{mvar|b}},{{mvar|c}}, {{mvar|d}}, {{mvar|e}} result in a [[Tamari lattice]] of order four, possibly different products.]]
If a binary operation is associative, repeated application of the operation produces the same result regardless of how valid pairs of parentheses are inserted in the expression.<ref>{{cite book |last=Durbin |first=John R. |title=Modern Algebra: an Introduction |year=1992 |publisher=Wiley |location=New York |isbn=978-0-471-51001-7 |page=78 |url=http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP000258.html |edition=3rd |quote=If <math>a_1, a_2, \dots, a_n \,\, (n \ge 2)</math> are elements of a set with an associative operation, then the product <math>a_1 a_2 \cdots a_n</math> is unambiguous; this is, the same element will be obtained regardless of how parentheses are inserted in the product.}}</ref> This is called the '''generalized associative law'''.

The number of possible bracketings is just the [[Catalan number]], <math>C_n</math>
, for ''n'' operations on ''n+1'' values. For instance, a product of 3 operations on 4 elements may be written (ignoring permutations of the arguments), in <math>C_3 = 5</math> possible ways:

*<math>((ab)c)d</math>
*<math>(a(bc))d</math>
*<math>a((bc)d)</math>
*<math>(a(b(cd))</math>
*<math>(ab)(cd)</math>

If the product operation is associative, the generalized associative law says that all these expressions will yield the same result. So unless the expression with omitted parentheses already has a different meaning (see below), the parentheses can be considered unnecessary and "the" product can be written unambiguously as
:<math>abcd</math>
As the number of elements increases, the [[Catalan number#Applications in combinatorics|number of possible ways to insert parentheses]] grows quickly, but they remain unnecessary for disambiguation.

An example where this does not work is the [[logical biconditional]] {{math|↔}}. It is associative; thus, {{math|{{var|A}} ↔ ({{var|B}} ↔ {{var|C}})}} is equivalent to {{math|({{var|A}} ↔ {{var|B}}) ↔ {{var|C}}}}, but {{math|{{var|A}} ↔ {{var|B}} ↔ {{var|C}}}} most commonly means {{math|({{var|A}} ↔ {{var|B}}) and ({{var|B}} ↔ {{var|C}})}}, which is not equivalent.