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These values allow logical conditions to be evaluated as either false or true."},{"text":"0 and 2 (or False and True)","correct":false,"explanation":"The False and True labels are appropriate, but Boolean numeric values are 0 and 1, not 0 and 2. Using 2 would introduce a third, invalid numeric state."},{"text":"1 and 2 (or False and True)","correct":false,"explanation":"Boolean variables use 0 and 1 as their numeric representations. Although False and True are the correct logical labels, replacing 0 with 2 gives an invalid Boolean value."},{"text":"0 and 1 (or Null and True)","correct":false,"explanation":"The numeric values are correct, but Null is not the Boolean counterpart of True. The two logical values are False and True, representing 0 and 1 respectively."}]},{"id":"55fcfc44-83e5-4d82-9263-32d5bc631553","cardType":"cloze","points":null,"front":"In Boolean algebra, the [ ] operator is applied first, followed by AND, and finally OR.","back":"In Boolean algebra, the NOT operator is applied first, followed by AND, and finally OR.","mcqOptions":[{"text":"NOT","correct":true,"explanation":"NOT has the highest precedence in Boolean algebra, so it is evaluated before AND and OR. It reverses a Boolean value, changing 1 to 0 or 0 to 1."},{"text":"NAND","correct":false,"explanation":"NAND is a Boolean operation, but it is not the operator given highest precedence in the standard order. NOT is applied first, followed by AND and then OR."},{"text":"AND","correct":false,"explanation":"AND is evaluated before OR, but it does not have the highest precedence. NOT is applied first in Boolean algebra because it operates directly on a single value."},{"text":"OR","correct":false,"explanation":"OR has the lowest precedence of these three operators, so it is applied last rather than first. The standard order is NOT, then AND, then OR."}]},{"id":"c124b695-3714-4562-ae25-57b6a98d4ba0","cardType":"cloze","points":null,"front":"In Boolean algebra, the NOT operator is applied first, followed by [ ], and finally OR.","back":"In Boolean algebra, the NOT operator is applied first, followed by AND, and finally OR.","mcqOptions":[{"text":"AND","correct":true,"explanation":"After NOT has been applied, AND is evaluated before OR in Boolean algebra. This precedence order ensures that expressions are interpreted consistently without requiring brackets."},{"text":"NAND","correct":false,"explanation":"NAND is not the standard second stage of Boolean precedence. The usual order is NOT first, AND second, and OR last, so the correct operator is AND."},{"text":"OR","correct":false,"explanation":"OR is evaluated after AND, not before it. The precedence sequence is NOT, followed by AND, and finally OR, making OR the last operator in the sentence."},{"text":"XOR","correct":false,"explanation":"XOR is a Boolean operator but is not the operator specified in the standard precedence sequence here. AND follows NOT, while OR is applied last."}]},{"id":"7551c6a2-71a2-40b0-8dfc-72a131187cdc","cardType":"cloze","points":null,"front":"In Boolean algebra, the NOT operator is applied first, followed by AND, and finally [ ].","back":"In Boolean algebra, the NOT operator is applied first, followed by AND, and finally OR.","mcqOptions":[{"text":"OR","correct":true,"explanation":"OR has the lowest precedence among NOT, AND, and OR, so it is applied last. The standard Boolean evaluation order is NOT, then AND, followed by OR."},{"text":"AND","correct":false,"explanation":"AND is evaluated before OR, so it cannot be the final operator in this precedence order. NOT comes first, AND second, and OR last."},{"text":"XOR","correct":false,"explanation":"XOR is a valid Boolean operation, but it is not the final operator in the stated standard precedence sequence. OR is applied after NOT and AND."},{"text":"NOT","correct":false,"explanation":"NOT has the highest precedence and is therefore applied first, not last. It is followed by AND, with OR having the lowest precedence."}]},{"id":"11c8798a-da10-46f3-8b89-ea8c672ba4f6","cardType":"basic","points":1,"front":"What does the symbol $\\oplus$ represent in Boolean notation?","back":"XOR (Exclusive OR)","mcqOptions":[{"text":"XOR (Exclusive OR)","correct":true,"explanation":"The symbol $\\oplus$ represents XOR, or Exclusive OR. Its output is 1 when exactly one input is 1, and 0 when both inputs have the same value."},{"text":"OR (Inclusive OR)","correct":false,"explanation":"The ordinary OR operator is usually written with $+$ and outputs 1 when either or both inputs are 1. The symbol $\\oplus$ specifically represents XOR, which excludes the both-true case."},{"text":"XNOR (Exclusive NOR)","correct":false,"explanation":"XNOR produces 1 when both inputs are equal, making it the inverse of XOR. The symbol $\\oplus$ represents XOR, which produces 1 when exactly one input is 1."},{"text":"NOR (NOT OR)","correct":false,"explanation":"NOR is the inversion of OR and produces 1 only when both inputs are 0. The symbol $\\oplus$ instead denotes XOR, which is true for different input values."}]},{"id":"d6ebd005-b730-4453-adbe-2bb88ba8cc40","cardType":"basic","points":1,"front":"Define equivalent Boolean expressions.","back":"Expressions that produce the same output for every possible combination of inputs.","mcqOptions":[{"text":"Expressions that produce the same output for every possible combination of inputs.","correct":true,"explanation":"Equivalent Boolean expressions may look different, but their truth tables match completely. For every possible combination of input values, both expressions produce the same output."},{"text":"Expressions that use the same operators for every possible combination of inputs.","correct":false,"explanation":"Equivalent expressions do not need to contain the same operators or be written in the same form. They are equivalent when their outputs match for every input combination."},{"text":"Expressions that produce the opposite output for every possible combination of inputs.","correct":false,"explanation":"Producing the opposite output describes inversion or complementation, not equivalence. Equivalent Boolean expressions must produce the same output for every possible combination of inputs."},{"text":"Expressions that produce the same output for one possible combination of inputs.","correct":false,"explanation":"Matching for only one input combination is insufficient to establish equivalence. Equivalent expressions must produce the same output for every possible combination of inputs."}]},{"id":"2298d3b6-0a9b-4331-96c7-9404456847f7","cardType":"basic","points":2,"front":"Why is simplifying a Boolean expression beneficial for hardware design?","back":"It can be implemented using fewer logic gates or fewer operations.","mcqOptions":[{"text":"It can be implemented using fewer logic gates or fewer operations.","correct":true,"explanation":"Simplification removes unnecessary Boolean operations, allowing the circuit to use fewer logic gates. This can reduce hardware cost, physical size, power consumption, and propagation delay."},{"text":"It can be implemented using more logic gates or more operations.","correct":false,"explanation":"Simplification aims to reduce, not increase, the required hardware. Using fewer logic gates or operations can lower cost and delay, so increasing them is the opposite effect."},{"text":"It can be implemented using the same logic gates or the same operations.","correct":false,"explanation":"A simplified expression may require fewer gates or operations than the original. Claiming the number always stays the same misses the main hardware-design benefit of simplification."},{"text":"It can be implemented using fewer input variables or more operations.","correct":false,"explanation":"Simplification primarily reduces gates or operations, not necessarily the number of input variables. The hardware benefit comes from implementing the same logic with fewer components or steps."}]},{"id":"19486822-244b-49f2-b0dd-977f1d6d6130","cardType":"basic","points":2,"front":"State the Null laws for OR and AND.","back":"

$A + 1 = 1$
$A \\cdot 0 = 0$

","mcqOptions":[{"text":"

$A + 1 = 1$
$A \\cdot 0 = 0$

","correct":true,"explanation":"The Null laws state that OR with 1 always produces 1, while AND with 0 always produces 0. These results are independent of the value of $A$."},{"text":"

$A + 0 = A$
$A \\cdot 1 = A$

","correct":false,"explanation":"These are the Identity laws, not the Null laws: OR with 0 and AND with 1 leave $A$ unchanged. The Null laws are $A + 1 = 1$ and $A \\cdot 0 = 0$."},{"text":"

$A + A = A$
$A \\cdot A = A$

","correct":false,"explanation":"These are the Idempotent laws, which describe combining a value with itself. The Null laws instead use constants that force the result: $A + 1 = 1$ and $A \\cdot 0 = 0$."},{"text":"

$A + 1 = A$
$A \\cdot 0 = A$

","correct":false,"explanation":"These equations incorrectly claim that the constants leave $A$ unchanged. In fact, OR with 1 gives 1 and AND with 0 gives 0, regardless of $A$."}]},{"id":"99d6fdb3-5a89-4734-82d2-fea7fe55ccdf","cardType":"basic","points":2,"front":"State the Identity laws for OR and AND.","back":"

$A + 0 = A$
$A \\cdot 1 = A$

","mcqOptions":[{"text":"

$A + 0 = A$
$A \\cdot 1 = A$

","correct":true,"explanation":"OR with 0 and AND with 1 both leave the original Boolean value unchanged. Therefore, 0 is the identity for OR and 1 is the identity for AND."},{"text":"

$A + 1 = A$
$A \\cdot 0 = A$

","correct":false,"explanation":"This reverses the identity values. OR with 1 always gives 1, while AND with 0 always gives 0; the identity laws use 0 for OR and 1 for AND."},{"text":"

$A + 0 = 0$
$A \\cdot 1 = 1$

","correct":false,"explanation":"This incorrectly treats the identity values as overriding values. Neither operation becomes constant: OR with 0 and AND with 1 both preserve A."},{"text":"

$A + A = A$
$A \\cdot A = A$

","correct":false,"explanation":"These are the idempotent laws, not the identity laws. The identity laws combine A with neutral constants: $A + 0 = A$ and $A \\cdot 1 = A$."}]},{"id":"1b77b2d3-a957-487d-9df7-dde79edcb00a","cardType":"basic","points":1,"front":"What is the result of the Idempotent laws ($A + A$ and $A \\cdot A$)?","back":"$$A$","mcqOptions":[{"text":"$$A$","correct":true,"explanation":"Combining a Boolean variable with itself does not change its value. Consequently, both $A + A$ and $A \\cdot A$ simplify to $A$."},{"text":"$$0$","correct":false,"explanation":"A repeated variable is not necessarily false; it retains its original value. The idempotent laws state that both $A + A$ and $A \\cdot A$ equal $A$."},{"text":"$$1$","correct":false,"explanation":"A repeated variable is not necessarily true; its value still depends on A. By the idempotent laws, both expressions simplify to $A$, not the constant 1."},{"text":"$$\\overline{A}$","correct":false,"explanation":"Repeating A does not complement it. Complementation requires the NOT operator, whereas the idempotent laws simplify both $A + A$ and $A \\cdot A$ directly to $A$."}]},{"id":"9d3d1db9-09de-4454-adae-1a6593c918af","cardType":"basic","points":2,"front":"State the two Complement laws.","back":"

$A + \\overline{A} = 1$
$A \\cdot \\overline{A} = 0$

","mcqOptions":[{"text":"

$A + \\overline{A} = 1$
$A \\cdot \\overline{A} = 0$

","correct":true,"explanation":"A and its complement must have opposite values. Therefore, their OR is always 1 because one is true, while their AND is always 0 because both cannot be true."},{"text":"

$A + \\overline{A} = 0$
$A \\cdot \\overline{A} = 1$

","correct":false,"explanation":"The outputs have been reversed. A and its complement include one true value, making their OR 1, but they can never both be true, making their AND 0."},{"text":"

$A + A = 1$
$A \\cdot A = 0$

","correct":false,"explanation":"This omits the complement and incorrectly assigns constant results. Complement laws require A and $\\overline{A}$, giving $A + \\overline{A} = 1$ and $A \\cdot \\overline{A} = 0$."},{"text":"

$A + \\overline{A} = A$
$A \\cdot \\overline{A} = A$

","correct":false,"explanation":"This confuses complement laws with laws that preserve A. Since A and its complement are opposites, their OR is always 1 and their AND is always 0."}]},{"id":"32b9ffe0-e79a-431c-9a4f-9485e6b638cc","cardType":"cloze","points":null,"front":"The Double Negation law states that $\\text{[...]}$.","back":"The Double Negation law states that $\\boxed{\\overline{\\overline{A}} = A}$.","mcqOptions":[{"text":"$$\\overline{\\overline{A}} = A$","correct":true,"explanation":"The first negation reverses A, and the second reverses it again. Two successive NOT operations therefore restore the original value, so double negation simplifies to A."},{"text":"$$\\overline{A} = A$","correct":false,"explanation":"A single negation normally changes the Boolean value rather than preserving it. Double negation specifically applies two NOT operations, giving $\\overline{\\overline{A}} = A$."},{"text":"$$\\overline{\\overline{A}} = \\overline{A}$","correct":false,"explanation":"This incorrectly treats the second negation as having no effect. It reverses the first negation, so two successive complements restore A rather than leaving its complement."},{"text":"$$\\overline{\\overline{A}} = 1$","correct":false,"explanation":"Double negation does not force a constant true result; it restores whatever value A originally held. Thus $\\overline{\\overline{A}}$ equals $A$, which may be 0 or 1."}]},{"id":"b51c2a14-2766-421a-9e1f-70b2219b6e1a","cardType":"basic","points":2,"front":"State the Distributive law for $A + (B \\cdot C)$.","back":"$$(A + B) \\cdot (A + C)$","mcqOptions":[{"text":"$$(A + B) \\cdot (A + C)$","correct":true,"explanation":"Boolean OR distributes over AND in this form. Expanding the right-hand side and applying absorption leaves precisely $A + (B \\cdot C)$."},{"text":"$$(A \\cdot B) + (A \\cdot C)$","correct":false,"explanation":"This is the distributive expansion of $A \\cdot (B + C)$, so it applies the wrong outer operator. For $A + (B \\cdot C)$, the correct form is $(A + B) \\cdot (A + C)$."},{"text":"$$(A + B) \\cdot C$","correct":false,"explanation":"This fails to combine A with C and therefore loses cases where A is true but C is false. Both factors are required: $(A + B) \\cdot (A + C)$."},{"text":"$$(A + B) + (A + C)$","correct":false,"explanation":"Using OR between the two brackets changes the expression to $A + B + C$, which is not equivalent. Distribution requires AND between the factors: $(A + B) \\cdot (A + C)$."}]},{"id":"e4b3cb1a-731c-4bab-b52e-95eba588eb51","cardType":"basic","points":2,"front":"State the two Absorption laws.","back":"

$A + A \\cdot B = A$
$A \\cdot (A + B) = A$

","mcqOptions":[{"text":"

$A + A \\cdot B = A$
$A \\cdot (A + B) = A$

","correct":true,"explanation":"In each expression, A determines the result regardless of B. The additional term cannot alter the outcome, so A absorbs $A \\cdot B$ in OR and $A + B$ in AND."},{"text":"

$A + A \\cdot B = B$
$A \\cdot (A + B) = B$

","correct":false,"explanation":"This incorrectly makes B determine the result. A is the repeated variable and absorbs the term involving B, so both expressions simplify to A."},{"text":"

$A + A \\cdot B = A + B$
$A \\cdot (A + B) = A \\cdot B$

","correct":false,"explanation":"This effectively discards one occurrence of A without applying absorption. The repeated A controls each expression, making both $A + A \\cdot B$ and $A \\cdot (A + B)$ simplify to A."},{"text":"

$A + A \\cdot B = A \\cdot B$
$A \\cdot (A + B) = A + B$

","correct":false,"explanation":"This swaps the expressions’ dominant operations rather than simplifying them. Under absorption, the extra B-dependent component has no effect, and each complete expression reduces to A."}]},{"id":"11d7ffdd-a6f8-4ef2-9029-289a2af0ccab","cardType":"cloze","points":null,"front":"According to De Morgan's Laws, $\\overline{A \\cdot B}$ is equivalent to $\\text{[...]}$.","back":"According to De Morgan's Laws, $\\overline{A \\cdot B}$ is equivalent to $\\boxed{\\overline{A} + \\overline{B}}$.","mcqOptions":[{"text":"$$\\overline{A} + \\overline{B}$","correct":true,"explanation":"De Morgan's Laws complement each variable and switch the operator. Negating an AND expression therefore produces an OR expression containing the complements of both A and B."},{"text":"$$\\overline{A} \\cdot \\overline{B}$","correct":false,"explanation":"This complements each variable but fails to switch AND to OR. De Morgan's Laws require both changes, so $\\overline{A \\cdot B}$ equals $\\overline{A} + \\overline{B}$."},{"text":"$$A + B$","correct":false,"explanation":"This switches AND to OR but incorrectly removes the negation from both variables. De Morgan's Laws give an OR of the individual complements: $\\overline{A} + \\overline{B}$."},{"text":"$$\\overline{A + B}$","correct":false,"explanation":"This changes the operator inside the original negation without distributing that negation. Applying De Morgan's Laws to $\\overline{A \\cdot B}$ gives $\\overline{A} + \\overline{B}$."}]},{"id":"1b5d0f88-0b90-4922-9469-68233cdae261","cardType":"cloze","points":null,"front":"According to De Morgan's Laws, $\\overline{A + B}$ is equivalent to $\\text{[...]}$.","back":"According to De Morgan's Laws, $\\overline{A + B}$ is equivalent to $\\boxed{\\overline{A} \\cdot \\overline{B}}$.","mcqOptions":[{"text":"$$\\overline{A} \\cdot \\overline{B}$","correct":true,"explanation":"De Morgan's Laws complement each variable and switch the operator. Negating an OR expression therefore produces an AND expression containing the complements of both A and B."},{"text":"$$\\overline{A} + \\overline{B}$","correct":false,"explanation":"This complements both variables but incorrectly retains OR. De Morgan's Laws also switch the operator, so $\\overline{A + B}$ equals $\\overline{A} \\cdot \\overline{B}$."},{"text":"$$A \\cdot B$","correct":false,"explanation":"This switches OR to AND but incorrectly removes the negation from both variables. De Morgan's Laws give an AND of the individual complements: $\\overline{A} \\cdot \\overline{B}$."},{"text":"$$\\overline{A \\cdot B}$","correct":false,"explanation":"This changes the operator inside the negation instead of distributing the negation across both variables. The correct equivalent is $\\overline{A} \\cdot \\overline{B}$."}]},{"id":"93704197-6f23-447b-b841-cc39091b03ea","cardType":"basic","points":2,"front":"What is the shortcut for applying De Morgan's law to a barred expression?","back":"Break the bar and change the operator (AND to OR, or OR to AND).","mcqOptions":[{"text":"Break the bar and change the operator (AND to OR, or OR to AND).","correct":true,"explanation":"De Morgan’s law distributes the negation across every term and swaps the Boolean operator: AND becomes OR, while OR becomes AND."},{"text":"Break the bar and keep the operator (AND stays AND, or OR stays OR).","correct":false,"explanation":"Keeping the operator is the common mistake here. De Morgan’s law requires the operator to change: AND becomes OR, and OR becomes AND."},{"text":"Keep the bar and change each variable (A to B, or B to A).","correct":false,"explanation":"Swapping variables does not apply De Morgan’s law. The bar must be distributed across the terms, and the operator must change between AND and OR."},{"text":"Remove the bar and reverse the variables (A then B, or B then A).","correct":false,"explanation":"Reversing variables only uses commutativity and does not handle the negation correctly. De Morgan’s law breaks the bar and changes AND to OR or vice versa."}]},{"id":"b981238a-90c3-460b-8b50-bf1dc9f4df09","cardType":"basic","points":2,"front":"Simplify the expression: $A \\cdot B + A \\cdot \\overline{B}$.","back":"$$A$","mcqOptions":[{"text":"$$A$","correct":true,"explanation":"Factorising gives $A(B+\\overline{B})$. Since $B+\\overline{B}=1$ by the complement law, the expression becomes $A\\cdot1=A$."},{"text":"$$B$","correct":false,"explanation":"Factoring out $B$ is invalid because the second term contains $\\overline{B}$. Factoring out $A$ gives $A(B+\\overline{B})=A\\cdot1=A$."},{"text":"$$A+B$","correct":false,"explanation":"The products cannot be changed directly into an OR expression. Factoring out $A$ and applying $B+\\overline{B}=1$ simplifies the original expression to $A$."},{"text":"$$A\\cdot B$","correct":false,"explanation":"This incorrectly discards the term containing $\\overline{B}$. Together, $B$ and $\\overline{B}$ cover every possible value, so the expression simplifies to $A$."}]},{"id":"031dcf2b-7146-471d-b14e-26e3b7e01b24","cardType":"basic","points":1,"front":"What method can be used to prove that a simplified expression is equivalent to the original?","back":"Compare the results of both using a truth table.","mcqOptions":[{"text":"Compare the results of both using a truth table.","correct":true,"explanation":"A truth table checks every possible combination of input values. If both expressions produce the same output in every row, they are logically equivalent."},{"text":"Compare the results of both using a single input.","correct":false,"explanation":"One matching input cannot establish equivalence because the outputs may differ for another combination. A truth table proves equivalence by checking every possible input combination."},{"text":"Compare the complexity of both using a gate count.","correct":false,"explanation":"Expressions with different numbers of gates can still be equivalent, while expressions with equal gate counts may differ. Their outputs must be compared using a truth table."},{"text":"Compare the written forms of both using a symbol count.","correct":false,"explanation":"The number of symbols indicates neither meaning nor equivalence. A truth table is needed to show that both expressions produce identical outputs for all possible inputs."}]},{"id":"bbf2af43-c8d4-486c-baeb-f1c239428384","cardType":"cloze","points":null,"front":"In Boolean algebra, $1 + 1 = \\text{[...]} because the $+$ symbol represents the OR operation.","back":"In Boolean algebra, $1 + 1 = \\boxed{1} because the $+$ symbol represents the OR operation.","mcqOptions":[{"text":"$$1$","correct":true,"explanation":"The $+$ symbol represents inclusive OR in Boolean algebra. Since OR produces $1$ when at least one input is $1$, $1+1$ equals $1$."},{"text":"$$0$","correct":false,"explanation":"This confuses OR with exclusive OR, for which two true inputs produce false. Boolean $+$ means inclusive OR, so $1+1$ equals $1$."},{"text":"$$2$","correct":false,"explanation":"This applies ordinary arithmetic addition instead of Boolean OR. Boolean values remain either $0$ or $1$, and $1$ OR $1$ produces $1$."},{"text":"$$10$","correct":false,"explanation":"This treats the expression as binary arithmetic, where one plus one is written $10$. The operator is Boolean OR here, so the result is $1$."}]},{"id":"000a5543-4cbd-4df1-8267-9d363a4eac56","cardType":"cloze","points":null,"front":"In Boolean algebra, $1 + 1 = 1 because the $+$ symbol represents the \\text{[...]} operation.","back":"In Boolean algebra, $1 + 1 = 1 because the $+$ symbol represents the \\boxed{OR} operation.","mcqOptions":[{"text":"OR","correct":true,"explanation":"In standard Boolean notation, $+$ denotes the OR operation. OR outputs $1$ whenever at least one input is $1$, including when both inputs are $1$."},{"text":"AND","correct":false,"explanation":"Although $1$ AND $1$ also gives $1$, AND is represented by a dot or adjacency, not $+$. The $+$ symbol represents OR."},{"text":"XOR","correct":false,"explanation":"XOR outputs $1$ only when exactly one input is $1$, so $1$ XOR $1$ would be $0$. Boolean $+$ represents inclusive OR."},{"text":"NOT","correct":false,"explanation":"NOT is a unary operation applied to one value and is normally shown with a bar or negation symbol. The binary $+$ operator represents OR."}]},{"id":"ecd9ad7c-7261-44f7-b330-283c60d9be60","cardType":"basic","points":1,"front":"Evaluate $F = \\overline{A} \\cdot B$ when $A=1$ and $B=1$.","back":"$$0$","mcqOptions":[{"text":"$$0$","correct":true,"explanation":"When $A=1$, its complement $\\overline{A}$ is $0$. Therefore $\\overline{A}\\cdot B=0\\cdot1=0$, because AND requires both inputs to be $1$."},{"text":"$$1$","correct":false,"explanation":"This overlooks the bar above $A$ and substitutes $A=1$ directly. Negating $A$ gives $0$, so $0$ AND $1$ evaluates to $0$."},{"text":"$$-1$","correct":false,"explanation":"This treats Boolean negation like arithmetic negation. The complement of Boolean $1$ is $0$, not $-1$, so the AND expression evaluates to $0$."},{"text":"$$2$","correct":false,"explanation":"Boolean operations produce only $0$ or $1$, so ordinary multiplication or addition cannot produce $2$. Here $\\overline{1}=0$, and $0\\cdot1=0$."}]},{"id":"8bcb17f0-958d-4e27-9975-b1df2882017c","cardType":"cloze","points":null,"front":"The Commutative Laws state that $A + B = \\text{[...]} and $A \\cdot B = $B \\cdot A$.","back":"The Commutative Laws state that $A + B = \\boxed{B + A} and $A \\cdot B = $B \\cdot A$.","mcqOptions":[{"text":"$$B + A$","correct":true,"explanation":"The commutative law allows the operands of OR to be exchanged without changing the result. Therefore $A+B$ is equivalent to $B+A$."},{"text":"$$A + B$","correct":false,"explanation":"Leaving the operands unchanged does not demonstrate the commutative rearrangement requested by the blank. Their order should be reversed while retaining OR, giving $B+A$."},{"text":"$$B \\cdot A$","correct":false,"explanation":"This reverses the operands but also changes OR to AND. Commutativity changes only operand order, so $A+B$ becomes $B+A$."},{"text":"$$\\overline{B + A}$","correct":false,"explanation":"Commutativity does not introduce a negation; it only reverses operand order. The equivalent form of $A+B$ is therefore $B+A$, without a bar."}]},{"id":"1f9bcd73-39de-465c-92d1-17aed3ca9d31","cardType":"cloze","points":null,"front":"The Commutative Laws state that $A + B = B + A and $A \\cdot B = $\\text{[...]}$.","back":"The Commutative Laws state that $A + B = B + A and $A \\cdot B = $\\boxed{B \\cdot A}$.","mcqOptions":[{"text":"$$B \\cdot A$","correct":true,"explanation":"The commutative law for AND permits the operands to be exchanged without affecting the output. Thus $A\\cdot B$ is equivalent to $B\\cdot A$."},{"text":"$$A \\cdot B$","correct":false,"explanation":"This repeats the original ordering rather than showing the commutative rearrangement. Reversing the operands while keeping the AND operator gives $B\\cdot A$."},{"text":"$$B + A$","correct":false,"explanation":"This reverses the operands but incorrectly changes AND to OR. The commutative law preserves the operator, so the correct rearrangement is $B\\cdot A$."},{"text":"$$\\overline{B \\cdot A}$","correct":false,"explanation":"Commutativity does not complement the expression; it only swaps the order of its operands. The equivalent AND expression is $B\\cdot A$, with no bar."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Paper 1"},"heading":"Boolean algebra","title":"Boolean algebra Flashcards","topbar":{"backHref":"/a-level/computer-science/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"e860b7c9-6b71-57ca-aff0-216bb3480ed4","walled":"$L38"}],"$L39","$L3a"]