[(𝑝 ∨ π‘ž) ∧ (𝑝 β†’ π‘Ÿ) ∧ (π‘ž β†’ π‘Ÿ)] β†’ π‘Ÿ ≑ T

asked by guest
on Mar 29, 2025 at 7:37 pm



MathBot Answer:

Logic Expression
Symbol: (((p∨q)∧(pβ†’r)∧(qβ†’r))β†’r)↔True\left(\left(\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r\right) \leftrightarrow \text{True}
Text: (((p or q) and (p implies r) and (q implies r)) implies r) if and only if True\left(\left(\left(p \text{ or } q\right) \text{ and } \left(p \text{ implies } r\right) \text{ and } \left(q \text{ implies } r\right)\right) \text{ implies } r\right) \text{ if and only if } \text{True}


Classification tautology


Truth Table

Legend

ppqqrrp∨qp \vee qpβ†’rp \rightarrow rqβ†’rq \rightarrow r(p∨q)∧(pβ†’r)∧(qβ†’r)\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)((p∨q)∧(pβ†’r)∧(qβ†’r))β†’r\left(\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r(((p∨q)∧(pβ†’r)∧(qβ†’r))β†’r)↔True\left(\left(\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r\right) \leftrightarrow \text{True}
111111111
110100011
101111111
100101011
011111111
010110011
001011011
000011011
T/F

Simplification(((p∨q)∧(pβ†’r)∧(qβ†’r))β†’r)↔True≑((p∨q)∧(pβ†’r)∧(qβ†’r))β†’rBiconditional Simplification≑((p∨q)∧(Β¬p∨r)∧(qβ†’r))β†’rConditional Equivalence≑((p∨q)∧(Β¬p∨r)∧(Β¬q∨r))β†’rConditional Equivalence≑((p∨q)∧(r∨(Β¬p∧¬q)))β†’rDistributive Law≑((p∨q)∧(r∨¬(p∨q)))β†’rDe Morgan’s Law≑((p∨q)∧r)β†’rRedundancy Law (2)≑TrueConjunctive Simplification\begin{gathered} \left(\left(\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r\right) \leftrightarrow \text{True} & \equiv & \left(\left(p \vee q\right) \wedge \left(p \rightarrow r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r & \text{Biconditional Simplification} \\ & \equiv & \left(\left(p \vee q\right) \wedge \left(\neg p \vee r\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow r & \text{Conditional Equivalence} \\ & \equiv & \left(\left(p \vee q\right) \wedge \left(\neg p \vee r\right) \wedge \left(\neg q \vee r\right)\right) \rightarrow r & \text{Conditional Equivalence} \\ & \equiv & \left(\left(p \vee q\right) \wedge \left(r \vee \left(\neg p \wedge \neg q\right)\right)\right) \rightarrow r & \text{Distributive Law} \\ & \equiv & \left(\left(p \vee q\right) \wedge \left(r \vee \neg \left(p \vee q\right)\right)\right) \rightarrow r & \text{De Morgan's Law} \\ & \equiv & \left(\left(p \vee q\right) \wedge r\right) \rightarrow r & \text{Redundancy Law (2)} \\ & \equiv & \text{True} & \text{Conjunctive Simplification} \end{gathered} Note: Solution may not be as simplified as possible.


Conjunctive Normal FormTrue\text{True}


Note 1: These equivalences and tautologies are used to generate the above steps.
Note 2: Two logical statements are logically equivalent if they always produce the same truth value. Consequently, p ≑ q is same as saying p ↔ q is a tautology. EquivalenceAbsorption Lawp∧(p∨q)≑pp∨(p∧q)≑pBiconditional Equivalencep↔q≑(p∨¬q)∧(Β¬p∨q)p↔q≑(p∧q)∨(Β¬p∧¬q)Biconditional Simplificationp↔p≑Truep↔True≑pp↔¬p≑Falsep↔False≑¬pComplement Lawp∧¬p≑Falsep∨¬p≑TrueConditional Equivalencepβ†’q≑¬p∨qConditional Simplificationpβ†’p≑Truepβ†’True≑Truepβ†’False≑¬ppβ†’Β¬p≑¬pTrueβ†’p≑pFalseβ†’p≑TrueΒ¬pβ†’p≑pConsensus Law(p∨q)∧(Β¬p∨r)∧(q∨r)≑(p∨q)∧(Β¬p∨r)(p∧q)∨(Β¬p∧r)∨(q∧r)≑(p∧q)∨(Β¬p∧r)De Morgan’s LawΒ¬(p∧q)≑¬p∨¬qΒ¬(p∨q)≑¬p∧¬qDistributive Lawp∧(q∨r)≑(p∧q)∨(p∧r)p∨(q∧r)≑(p∨q)∧(p∨r)(p∨q)∧(r∨s)≑(p∧r)∨(p∧s)∨(q∧r)∨(q∧s)(p∧q)∨(r∧s)≑(p∨r)∧(p∨s)∧(q∨r)∧(q∨s)Domination Lawp∨True≑Truep∧False≑FalseDouble Negation LawΒ¬(Β¬p)≑pIdempotent Lawp∧p≑pp∨p≑pIdentity Lawp∧True≑pp∨False≑pNANDp↑q≑¬(p∧q)Negation LawΒ¬True≑FalseΒ¬False≑TrueNORp↓q≑¬(p∨q)Negation of Biconditional EquivalenceΒ¬(p↔q)≑(p∨q)∧(Β¬p∨¬q)Β¬(p↔q)≑(p∧¬q)∨(Β¬p∧q)Negation of Conditional EquivalenceΒ¬(pβ†’q)≑p∧¬qRedundancy Law (1)(p∨q)∧(p∨¬q)≑p(p∧q)∨(p∧¬q)≑pRedundancy Law (2)p∧(Β¬p∨q)≑p∧qp∨(Β¬p∧q)≑p∨qXORpβŠ•q≑(p∨q)∧(Β¬p∨¬q)pβŠ•q≑(p∧¬q)∨(Β¬p∧q)XOR SimplificationpβŠ•p≑FalsepβŠ•True≑¬ppβŠ•Β¬p≑TruepβŠ•False≑pXNORpβŠ™q≑¬(pβŠ•q)\begin{array}{c|c}\textbf{Equivalence} \\ \hline \text{Absorption Law} & \begin{gathered} p \wedge \left(p \vee q\right) \equiv p \\ p \vee \left(p \wedge q\right) \equiv p \end{gathered} \\ \hline \text{Biconditional Equivalence} & \begin{gathered} p \leftrightarrow q \equiv \left(p \vee \neg q\right) \wedge \left(\neg p \vee q\right) \\ p \leftrightarrow q \equiv \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right) \end{gathered} \\ \hline \text{Biconditional Simplification} & \begin{gathered} p \leftrightarrow p \equiv \text{True} & p \leftrightarrow \text{True} \equiv p \\ p \leftrightarrow \neg p \equiv \text{False} & p \leftrightarrow \text{False} \equiv \neg p \end{gathered} \\ \hline \text{Complement Law} & \begin{gathered} p \wedge \neg p \equiv \text{False} \\ p \vee \neg p \equiv \text{True} \end{gathered} \\ \hline \text{Conditional Equivalence} & p \rightarrow q \equiv \neg p \vee q \\ \hline \text{Conditional Simplification} & \begin{gathered} p \rightarrow p \equiv \text{True} & p \rightarrow \text{True} \equiv \text{True} & p \rightarrow \text{False} \equiv \neg p \\ p \rightarrow \neg p \equiv \neg p & \text{True} \rightarrow p \equiv p & \text{False} \rightarrow p \equiv \text{True} \\ \neg p \rightarrow p \equiv p\end{gathered} \\ \hline \text{Consensus Law} & \begin{gathered} \left(p \vee q\right) \wedge \left(\neg p \vee r\right) \wedge \left(q \vee r\right) \equiv \left(p \vee q\right) \wedge \left(\neg p \vee r\right) \\ \left(p \wedge q\right) \vee \left(\neg p \wedge r\right) \vee \left(q \wedge r\right) \equiv \left(p \wedge q\right) \vee \left(\neg p \wedge r\right) \end{gathered} \\ \hline \text{De Morgan's Law} & \begin{gathered} \neg \left(p \wedge q\right) \equiv \neg p \vee \neg q \\ \neg \left(p \vee q\right) \equiv \neg p \wedge \neg q \end{gathered} \\ \hline \text{Distributive Law} & \begin{gathered} p \wedge \left(q \vee r\right) \equiv \left(p \wedge q\right) \vee \left(p \wedge r\right) \\ p \vee \left(q \wedge r\right) \equiv \left(p \vee q\right) \wedge \left(p \vee r\right) \\ \left(p \vee q\right) \wedge \left(r \vee s\right) \equiv \left(p \wedge r\right) \vee \left(p \wedge s\right) \vee \left(q \wedge r\right) \vee \left(q \wedge s\right) \\ \left(p \wedge q\right) \vee \left(r \wedge s\right) \equiv \left(p \vee r\right) \wedge \left(p \vee s\right) \wedge \left(q \vee r\right) \wedge \left(q \vee s\right) \end{gathered} \\ \hline \text{Domination Law} & \begin{gathered} p \vee \text{True} \equiv \text{True} \\ p \wedge \text{False} \equiv \text{False} \end{gathered} \\ \hline \text{Double Negation Law} & \neg \left(\neg p\right) \equiv p \\ \hline \text{Idempotent Law} & \begin{gathered} p \wedge p \equiv p \\ p \vee p \equiv p \end{gathered} \\ \hline \text{Identity Law} & \begin{gathered} p \wedge \text{True} \equiv p \\ p \vee \text{False} \equiv p \end{gathered} \\ \hline \text{NAND} & p \uparrow q \equiv \neg \left(p \wedge q\right) \\ \hline \text{Negation Law} & \begin{gathered} \neg \text{True} \equiv \text{False} \\ \neg \text{False} \equiv \text{True} \end{gathered} \\ \hline \text{NOR} & p \downarrow q \equiv \neg \left(p \vee q\right) \\ \hline \text{Negation of Biconditional Equivalence} & \begin{gathered} \neg \left(p \leftrightarrow q\right) \equiv \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right) \\ \neg \left(p \leftrightarrow q\right) \equiv \left(p \wedge \neg q\right) \vee \left(\neg p \wedge q\right) \end{gathered} \\ \hline \text{Negation of Conditional Equivalence} & \neg \left(p \rightarrow q\right) \equiv p \wedge \neg q \\ \hline \text{Redundancy Law (1)} & \begin{gathered} \left(p \vee q\right) \wedge \left(p \vee \neg q\right) \equiv p \\ \left(p \wedge q\right) \vee \left(p \wedge \neg q\right) \equiv p \end{gathered} \\ \hline \text{Redundancy Law (2)} & \begin{gathered} p \wedge \left(\neg p \vee q\right) \equiv p \wedge q \\ p \vee \left(\neg p \wedge q\right) \equiv p \vee q \end{gathered} \\ \hline \text{XOR} & \begin{gathered} p \oplus q \equiv \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right) \\ p \oplus q \equiv \left(p \wedge \neg q\right) \vee \left(\neg p \wedge q\right) \end{gathered} \\ \hline \text{XOR Simplification} & \begin{gathered} p \oplus p \equiv \text{False} & p \oplus \text{True} \equiv \neg p \\ p \oplus \neg p \equiv \text{True} & p \oplus \text{False} \equiv p \end{gathered} \\ \hline \text{XNOR} & p \odot q \equiv \neg \left(p \oplus q\right) \end{array} TautologyConjunctive Simplification(p∧q)β†’p(p∧q)β†’qContradictionΒ¬(p∧¬p)Contrapositive(pβ†’q)↔(Β¬qβ†’Β¬p)Disjunctive Additionpβ†’(p∨q)qβ†’(p∨q)Disjunctive Syllogism((p∨q)∧¬q)β†’p((p∨q)∧¬p)β†’qHypothetical Syllogism((pβ†’q)∧(qβ†’r))β†’(pβ†’r)Modus Ponens(p∧(pβ†’q))β†’qModus Tollens(Β¬q∧(pβ†’q))β†’Β¬p\begin{array}{c|c}\textbf{Tautology} \\ \hline \text{Conjunctive Simplification} & \begin{gathered} \left(p \wedge q\right) \rightarrow p \\ \left(p \wedge q\right) \rightarrow q \end{gathered} \\ \hline \text{Contradiction} & \neg \left(p \wedge \neg p\right) \\ \hline \text{Contrapositive} & \left(p \rightarrow q\right) \leftrightarrow \left(\neg q \rightarrow \neg p\right) \\ \hline \text{Disjunctive Addition} & \begin{gathered} p \rightarrow \left(p \vee q\right) \\ q \rightarrow \left(p \vee q\right) \end{gathered} \\ \hline \text{Disjunctive Syllogism} & \begin{gathered} \left(\left(p \vee q\right) \wedge \neg q\right) \rightarrow p \\ \left(\left(p \vee q\right) \wedge \neg p\right) \rightarrow q \end{gathered} \\ \hline \text{Hypothetical Syllogism} & \left(\left(p \rightarrow q\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow \left(p \rightarrow r\right) \\ \hline \text{Modus Ponens} & \left(p \wedge \left(p \rightarrow q\right)\right) \rightarrow q \\ \hline \text{Modus Tollens} & \left(\neg q \wedge \left(p \rightarrow q\right)\right) \rightarrow \neg p \end{array}