Boolean simplifier

Name: Boolean simplifier
Author: sajith tiyenshan

Android app by sajith tiyenshan. Education · sajith tiyenshan

Store rating: Unknown
Store rating count: Unknown
Download price: Free to download
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Version: 1.0
Listing last refreshed: 2026-09-13

Store description excerpt

this is web view app of "https://www.boolean-algebra.com" Boolean Postulate, Properties, and Theorems The following postulate, properties, and theorems are valid in Boolean Algebra and are used in simplification of logical expressions or functions: POSTULATES are self - evident truths. 1a: $A=1$ (if A ≠ 0) 1b: $A=0$ (if A ≠ 1) 2a: $0∙0=0$ 2b: $0+0=0$ 3a: $1∙1=1$ 3b: $1+1=1$ 4a: $1∙0=0$ 4b: $1+0=1$ 5a: $\overline{1}=0$ 5b: $\overline{0}=1$ PROPERTIES that are valid in Boolean Algebra are similar to the ones in ordinary algebra Commutative $A∙B=B∙A$ $A+B=B+A$ Associative $A∙(B∙C)=(A∙B)∙C$ $A+(B+C)=(A+B)+C$ Distributive $A∙(B+C)=A∙B+A∙C$ $A+(B∙C)=(A+B)∙(A+C)$ THEOREMS that are defined in Boolean Algebra are the following: 1a: $A∙0=0$ 1b: $A+0=A$ 2a: $A∙1=A$ 2b: $A+1=1$ 3a: $A∙A=A$ 3b: $A+A=A$ 4a: $A∙\overline{A}=0$ 4b: $A+\overline{A}=1$ 5a: $\overline{\overline{A}}=A$ 5b: $A=\overline{\overline{A}}$ 6a: $\overline{A∙B}=\overline{A}+\overline{B}$ 6b: $\overline{A+B}=\overline{A}∙\overline{B}$ By applying Boolean postulates, properties and/or theorems we can simplify complex Boolean expressions and build a smaller logic block diagram (less expensive circuit). For example, to simplify $AB(A+C)$ we have: $AB(A+C)$ distributive law =$ABA+ABC$ cumulative law =$AAB+ABC$ theorem 3a =$AB+ABC$ distributive law =$AB(1+C)$ theorem 2b =$AB1$ theorem 2a =$AB$ Although the above is all you need to simplify a Boolean equation. You can use an extension of the theorems/laws to make it easier to simplify. The following will reduce the amount of steps required to simplify but will be more difficult to identify. 7a: $A∙(A+B)=A$ 7b: $A+A∙B=A$ 8a: $(A+B)∙(A+\overline{B})=A$ 8b: $A∙B+A∙\overline{B}=A$ 9a: $(A+\overline{B})∙B=A∙B$ 9b: $A∙\overline{B}+B=A+B$ 10: $A⊕B=\overline{A}∙B+A∙\overline{B}$ 11: $A⊙B=\overline{A}∙\overline{B}+A∙B$ ⊕ = XOR, ⊙ = XNOR Now using these new theorems/laws we can simplify the previous expression like this. To simplify $AB(A+C)$ we have: $AB(A+C)$ distributive law =$ABA+ABC$ cumulative law =$AAB+ABC$ theorem 3a =$AB+ABC$ theorem 7b

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