Let-A-B-and-C-are-three-statments-A-B-C-A-C-B-A-B-C-A-B-A-Prove-or-disprove-

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Question Number 1656 by Rasheed Soomro last updated on 29/Aug/15

$$\mathrm{Let}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{three}\mathrm{statments}.$$$$(\mathrm{A}\Rightarrow \mathrm{B}\Rightarrow \mathrm{C}\Rightarrow \mathrm{A})\stackrel{?}{\Rightarrow }(\mathrm{C}\Rightarrow \mathrm{B})$$$$(\mathrm{A}\Rightarrow \mathrm{B}\Rightarrow \mathrm{C}\Rightarrow \mathrm{A})\stackrel{?}{\Rightarrow }(\mathrm{B}\Rightarrow \mathrm{A})$$$$\mathrm{Prove}\mathrm{or}\mathrm{disprove}.$$

Commented by Yozzian last updated on 29/Aug/15

$$Thebestapproachtoshowing$$$$whetherornotthosestatements$$$$aretautologiesorcontradictions$$$$isbyuseoftruthtables.$$

Commented by 112358 last updated on 29/Aug/15

$$A\Rightarrow B\Rightarrow C\Rightarrow A\stackrel{?}{\equiv }((A\Rightarrow B)\Rightarrow C)\Rightarrow A$$

Commented by Yozzy last updated on 29/Aug/15

$$Letp=(A\Rightarrow B\Rightarrow C\Rightarrow A)\Rightarrow (C\Rightarrow B)$$$$r=(A\Rightarrow B\Rightarrow C\Rightarrow A)\Rightarrow (B\Rightarrow A)$$$$I^{\prime }massumingthatA\Rightarrow B\Rightarrow C\Rightarrow A\equiv ((A\Rightarrow B)\Rightarrow C)\Rightarrow A\equiv (A\Rightarrow B)\Rightarrow (C\Rightarrow A).$$$$Letq=(A\Rightarrow B)\Rightarrow (C\Rightarrow A)$$$$123456789101112\left(columnnumbers\right)$$$$A\mid B\mid C\mid A\Rightarrow B\mid (A\Rightarrow B)\Rightarrow C\mid ((A\Rightarrow B)\Rightarrow C)\Rightarrow A\mid C\Rightarrow B\mid p\mid C\Rightarrow A\mid q\mid B\Rightarrow A\mid r$$$$0\mid 0\mid 0101111111$$$$0\mid 0\mid 1110010011$$$$0\mid 1\mid 0101111100$$$$0\mid 1\mid 1110110001$$$$1\mid 0\mid 0011111111$$$$1\mid 0\mid 1011001111$$$$1\mid 1\mid 0101111111$$$$1\mid 1\mid 1111111111$$$$(1indicatesastatementistruewhile0indicatesastatementisfalse.)$$$$Fromtheabovetruthtablecolumn8,wehavethatp=(A\Rightarrow B\Rightarrow C\Rightarrow A)\Rightarrow (C\Rightarrow B)\equiv 1ifwe$$$$haveanylogiccombinationofA,BandCotherthanA=1,B=0,C=1.$$$$ByBooleanalgebra,$$$$p=(1\Rightarrow 0\Rightarrow 1\Rightarrow 1)\Rightarrow (1\Rightarrow 0)$$$$=(0\Rightarrow 1\Rightarrow 1)\Rightarrow 0$$$$=(1\Rightarrow 1)\Rightarrow 0$$$$=\backsim (1\Rightarrow 1)\vee 0$$$$=\backsim 1\vee 0$$$$=0\vee 0$$$$p=0$$$$(A\Rightarrow B\Rightarrow C\Rightarrow \Rightarrow A)\nRightarrow (C\Rightarrow B)ifAistrue,BisfalseandCistrue.$$$$Also,column12showsthat(A\Rightarrow B\Rightarrow C\Rightarrow A)\Rightarrow (B\Rightarrow A)isfalseiffA=0,$$$$B=1andC=0.$$

Commented by Rasheed Soomro last updated on 30/Aug/15

$$\mathbf{Thanks}\mathbf{for}\mathbf{so}\mathbf{much}\mathbf{labour}.$$$$\mathrm{I}\mathrm{have}\mathrm{not}\mathrm{been}\mathrm{much}\mathrm{involved}\mathrm{in}\mathrm{logic}.$$$$\mathrm{So}\mathrm{if}\mathrm{you}\mathrm{see}\mathrm{no}\mathrm{meaning}\mathrm{in}\mathrm{my}\mathrm{questions}\mathrm{please}\mathrm{excuse}$$$$\mathrm{me}.$$$$\mathrm{Is}\mathrm{there}\mathrm{a}\mathrm{transitive}\mathrm{law}\mathrm{of}\mathrm{implication}?$$$$\mathrm{A}\Rightarrow \mathrm{B},\mathrm{B}\Rightarrow \mathrm{C}\mathrm{then}\mathrm{A}\Rightarrow \mathrm{C}?$$$${}^{\prime }\mathrm{If}\mathrm{A}\mathrm{is}\mathrm{true}\mathrm{then}\mathrm{B}\mathrm{is}{\mathrm{true}}^{\prime }{,}^{\prime }\mathrm{If}\mathrm{B}\mathrm{is}\mathrm{true}\mathrm{then}\mathrm{C}\mathrm{is}{\mathrm{true}}^{\prime }$$$$\Rightarrow$$$${}^{\prime }\mathrm{If}\mathrm{A}\mathrm{is}\mathrm{true}\mathrm{then}\mathrm{C}\mathrm{is}{\mathrm{true}}^{\prime }?$$$$\mathrm{Can}\mathrm{we}\mathrm{write}\mathrm{A}\Rightarrow \mathrm{B}\Rightarrow \mathrm{C}\Rightarrow \mathrm{A}\mathrm{in}\mathrm{place}\mathrm{of}$$$$\mathrm{A}\Rightarrow \mathrm{B},\mathrm{B}\Rightarrow \mathrm{C},\mathrm{C}\Rightarrow \mathrm{A}?$$

Commented by Yozzy last updated on 30/Aug/15

$$Proposethatp=A\Rightarrow B,q=B\Rightarrow Candr=A\Rightarrow C.$$$$Toprovetherebeingatransitivepropertyofimplicationislikesaying\left(Ithink\right)to$$$$showthat[(A\Rightarrow B)\wedge (B\Rightarrow C)]\Rightarrow (A\Rightarrow C)=T,i.e(p\wedge q)\Rightarrow risatautology.$$$$Withouttruthtables,I^{\prime }llattempttousethelawsofBooleanalgebra$$$$tosimplifythelogicexpressiontohopefullyenduphavingatautology.$$$$Notethat\mu \Rightarrow z\equiv \backsim \mu \vee z.$$$$[(A\Rightarrow B)\wedge (B\Rightarrow C)]\Rightarrow (A\Rightarrow C)\equiv [(\backsim A\vee B)\wedge (\backsim B\vee C)]\Rightarrow (\backsim A\vee C)$$$$lhs\equiv \backsim [(\backsim A\vee B)\wedge (\backsim B\vee C)]\vee (\backsim A\vee C)$$$$lhs\equiv [\backsim (\backsim A\vee B)\vee \backsim (\backsim B\vee C)]\vee (\backsim A\vee C)De{Morgan}^{\prime }slaw$$$$lhs\equiv [(A\wedge \backsim B)\vee (B\wedge \backsim C)]\vee (\backsim A\vee C)De{Morgan}^{\prime }slaw,Doublenegationlaw$$$$lhs\equiv [\backsim A\vee (A\wedge \backsim B)]\vee [C\vee (B\wedge \backsim C)]Commutativelaw,Associativelaw$$$$lhs\equiv [(\backsim A\vee A)\wedge (\backsim A\vee \backsim B)]\vee [(C\vee B)\wedge (C\vee \backsim C)]Distributionlaw$$$$lhs\equiv [T\wedge (\backsim A\vee \backsim B)]\vee [(C\vee B)\wedge T]Complementarylaw$$$$lhs\equiv (\backsim A\vee \backsim B)\vee (C\vee B)Identitylaw$$$$lhs\equiv (\backsim B\vee B)\vee (C\vee \backsim A)Commutativelaw,Associativelaw$$$$lhs\equiv T\vee (C\vee \backsim A)Complementarylaw$$$$lhs\equiv TDominationlaw$$$$ThesimplificationoftheBooleanstatementhasgivenusaconstant$$$$truthvalueofT,sothestatementisperpetuallytrue.Hence,atransitive$$$$propertyoftheimplicationlogicasbeenproven.\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$$$$$$

Commented by Yozzy last updated on 30/Aug/15

$$A\mid B\mid C\mid A\Rightarrow B\mid B\Rightarrow C\mid A\Rightarrow C\mid (A\Rightarrow B)\wedge (B\Rightarrow C)\mid [(A\Rightarrow B)\wedge (B\Rightarrow C)]\Rightarrow (A\Rightarrow C)$$$$00011111$$$$00111111$$$$01010101$$$$01111111$$$$10001001$$$$10101101$$$$11010001$$$$11111111$$$$Thelastcolumntotherightindicatesthatthestatement$$$$[(A\Rightarrow B)\wedge (B\Rightarrow C)]\Rightarrow (A\Rightarrow C)isalwaystrue.Therefore,thisproves$$$$theexistanceofatransitivepropertyoftheimplicationlogic.$$$$(I^{\prime }llcheckovermyBooleansimplification.Checked.)$$

Commented by Rasheed Soomro last updated on 30/Aug/15

$$\mathbf{THANKS}\mathrm{Ss}ssssss!$$$$\mathrm{Yozzian}\stackrel{?}{=}\mathrm{Yozzy}$$

Commented by Yozzy last updated on 30/Aug/15

$$Iusemycellphones\left(2\right)andmytablet.I^{\prime }vebeentheonlyonecommenting$$$$onthisquestion.Iusedmytablettoanswerthequestionbecauseofthe$$$$longtruthtable.$$

Commented by Rasheed Soomro last updated on 30/Aug/15

$$IhavealsomorethanonedeviceandIusetwo{ID}^{\prime }s:$$$$\mathrm{Rasheed}\mathrm{Soomro}and\mathrm{Rasheed}\mathrm{Ahmad}$$

Commented by 112358 last updated on 30/Aug/15

$$Isee.$$

Commented by 123456 last updated on 30/Aug/15

$$\mathrm{XD}{2}^{n_i}$$

Commented by Rasheed Ahmad last updated on 30/Aug/15

$$\mathrm{XD}{2}^{n_i}???$$

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