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

Let A,B and C are three statments. (A⇒B⇒C⇒A) ⇒^(?) (C⇒B) (A⇒B⇒C⇒A) ⇒^(?) (B⇒A) Prove or disprove.

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

Commented by Yozzian last updated on 29/Aug/15

The best approach to showing whether or not those statements are tautologies or contradictions is by use of truth tables.

$${The}\:{best}\:{approach}\:{to}\:{showing} \\ $$$${whether}\:{or}\:{not}\:{those}\:{statements} \\ $$$${are}\:{tautologies}\:{or}\:{contradictions} \\ $$$${is}\:{by}\:{use}\:{of}\:{truth}\:{tables}. \\ $$

Commented by 112358 last updated on 29/Aug/15

A⇒B⇒C⇒A≡^? ((A⇒B)⇒C)⇒A

$${A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\overset{?} {\equiv}\left(\left({A}\Rightarrow{B}\right)\Rightarrow{C}\right)\Rightarrow{A} \\ $$

Commented by Yozzy last updated on 29/Aug/15

Let p=(A⇒B⇒C⇒A)⇒(C⇒B) r=(A⇒B⇒C⇒A)⇒(B⇒A) I′m assuming that A⇒B⇒C⇒A≡((A⇒B)⇒C)⇒A≡(A⇒B)⇒(C⇒A) . Let q=(A⇒B)⇒(C⇒A) 1 2 3 4 5 6 7 8 9 10 11 12 (column numbers) A∣B∣C∣A⇒B∣(A⇒B)⇒C∣((A⇒B)⇒C)⇒A∣ C⇒B∣ p∣C⇒A∣q ∣B⇒A∣ r 0∣ 0 ∣0 1 0 1 1 1 1 1 1 1 0∣ 0 ∣1 1 1 0 0 1 0 0 1 1 0∣ 1 ∣0 1 0 1 1 1 1 1 0 0 0∣ 1 ∣1 1 1 0 1 1 0 0 0 1 1∣ 0 ∣0 0 1 1 1 1 1 1 1 1 1∣ 0 ∣1 0 1 1 0 0 1 1 1 1 1∣ 1 ∣0 1 0 1 1 1 1 1 1 1 1∣ 1 ∣1 1 1 1 1 1 1 1 1 1 (1 indicates a statement is true while 0 indicates a statement is false.) From the above truth table column 8, we have that p=(A⇒B⇒C⇒A)⇒(C⇒B)≡1 if we have any logic combination of A,B and C other than A=1,B=0,C=1. By Boolean algebra, p=(1⇒0⇒1⇒1)⇒(1⇒0) =(0⇒1⇒1)⇒0 =(1⇒1)⇒0 =∽(1⇒1)∨0 =∽1∨0 =0∨0 p=0 (A⇒B⇒C⇒⇒A)⇏(C⇒B) if A is true,B is false and C is true. Also, column 12 shows that (A⇒B⇒C⇒A)⇒(B⇒A) is false iff A=0 , B=1 and C=0.

$${Let}\:{p}=\left({A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\right)\Rightarrow\left({C}\Rightarrow{B}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:{r}=\left({A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\right)\Rightarrow\left({B}\Rightarrow{A}\right) \\ $$$${I}'{m}\:{assuming}\:{that}\:{A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\equiv\left(\left({A}\Rightarrow{B}\right)\Rightarrow{C}\right)\Rightarrow{A}\equiv\left({A}\Rightarrow{B}\right)\Rightarrow\left({C}\Rightarrow{A}\right)\:. \\ $$$${Let}\:{q}=\left({A}\Rightarrow{B}\right)\Rightarrow\left({C}\Rightarrow{A}\right)\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{1}\:\:\:\mathrm{2}\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{9}\:\:\:\:\:\:\:\:\mathrm{10}\:\:\:\:\:\:\mathrm{11}\:\:\:\:\mathrm{12}\:\left({column}\:{numbers}\right) \\ $$$${A}\mid{B}\mid{C}\mid{A}\Rightarrow{B}\mid\left({A}\Rightarrow{B}\right)\Rightarrow{C}\mid\left(\left({A}\Rightarrow{B}\right)\Rightarrow{C}\right)\Rightarrow{A}\mid\:{C}\Rightarrow{B}\mid\:{p}\mid{C}\Rightarrow{A}\mid{q}\:\:\mid{B}\Rightarrow{A}\mid\:{r} \\ $$$$\mathrm{0}\mid\:\:\mathrm{0}\:\mid\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{0}\mid\:\:\mathrm{0}\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{0}\mid\:\:\mathrm{1}\:\mid\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{0} \\ $$$$\mathrm{0}\mid\:\:\mathrm{1}\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\mid\:\:\mathrm{0}\:\mid\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\mid\:\:\mathrm{0}\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\mid\:\:\mathrm{1}\:\mid\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\mid\:\:\mathrm{1}\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\left(\mathrm{1}\:{indicates}\:{a}\:{statement}\:{is}\:{true}\:{while}\:\mathrm{0}\:{indicates}\:{a}\:{statement}\:{is}\:{false}.\right) \\ $$$${From}\:{the}\:{above}\:{truth}\:{table}\:{column}\:\mathrm{8},\:{we}\:{have}\:{that}\:{p}=\left({A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\right)\Rightarrow\left({C}\Rightarrow{B}\right)\equiv\mathrm{1}\:{if}\:{we} \\ $$$${have}\:{any}\:{logic}\:{combination}\:{of}\:{A},{B}\:{and}\:{C}\:{other}\:{than}\:{A}=\mathrm{1},{B}=\mathrm{0},{C}=\mathrm{1}. \\ $$$${By}\:{Boolean}\:{algebra},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{p}=\left(\mathrm{1}\Rightarrow\mathrm{0}\Rightarrow\mathrm{1}\Rightarrow\mathrm{1}\right)\Rightarrow\left(\mathrm{1}\Rightarrow\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{0}\Rightarrow\mathrm{1}\Rightarrow\mathrm{1}\right)\Rightarrow\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}\Rightarrow\mathrm{1}\right)\Rightarrow\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\backsim\left(\mathrm{1}\Rightarrow\mathrm{1}\right)\vee\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\backsim\mathrm{1}\vee\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}\vee\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{p}=\mathrm{0} \\ $$$$\left({A}\Rightarrow{B}\Rightarrow{C}\Rightarrow\Rightarrow{A}\right)\nRightarrow\left({C}\Rightarrow{B}\right)\:{if}\:{A}\:{is}\:{true},{B}\:{is}\:{false}\:{and}\:{C}\:{is}\:{true}. \\ $$$${Also},\:{column}\:\mathrm{12}\:{shows}\:{that}\:\left({A}\Rightarrow{B}\Rightarrow{C}\Rightarrow{A}\right)\Rightarrow\left({B}\Rightarrow{A}\right)\:{is}\:{false}\:{iff}\:{A}=\mathrm{0}\:, \\ $$$${B}=\mathrm{1}\:{and}\:{C}=\mathrm{0}. \\ $$

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

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

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

Commented by Yozzy last updated on 30/Aug/15

Propose that p=A⇒B,q=B⇒C and r=A⇒C . To prove there being a transitive property of implication is like saying (I think) to show that [(A⇒B)∧(B⇒C)]⇒(A⇒C)=T, i.e (p∧q)⇒r is a tautology. Without truth tables, I′ll attempt to use the laws of Boolean algebra to simplify the logic expression to hopefully end up having a tautology. Note that μ⇒z≡∽μ∨z. [(A⇒B)∧(B⇒C)]⇒(A⇒C)≡[(∽A∨B)∧(∽B∨C)]⇒(∽A∨C) lhs≡∽[(∽A∨B)∧(∽B∨C)]∨(∽A∨C) lhs≡[∽(∽A∨B)∨∽(∽B∨C)]∨(∽A∨C) De Morgan′s law lhs≡[(A∧∽B)∨(B∧∽C)]∨(∽A∨C) De Morgan′s law, Double negation law lhs≡[∽A∨(A∧∽B)]∨[C∨(B∧∽C)] Commutative law,Associative law lhs≡[(∽A∨A)∧(∽A∨∽B)]∨[(C∨B)∧(C∨∽C)] Distribution law lhs≡[T∧(∽A∨∽B)]∨[(C∨B)∧T] Complementary law lhs≡(∽A∨∽B)∨(C∨B) Identity law lhs≡(∽B∨B)∨(C∨∽A) Commutative law,Associative law lhs≡T∨(C∨∽A) Complementary law lhs≡T Domination law The simplification of the Boolean statement has given us a constant truth value of T, so the statement is perpetually true. Hence,a transitive property of the implication logic as been proven. ■

$${Propose}\:{that}\:{p}={A}\Rightarrow{B},{q}={B}\Rightarrow{C}\:{and}\:{r}={A}\Rightarrow{C}\:. \\ $$$${To}\:{prove}\:{there}\:{being}\:{a}\:{transitive}\:{property}\:{of}\:{implication}\:{is}\:{like}\:{saying}\:\left({I}\:{think}\right)\:{to}\: \\ $$$${show}\:{that}\:\left[\left({A}\Rightarrow{B}\right)\wedge\left({B}\Rightarrow{C}\right)\right]\Rightarrow\left({A}\Rightarrow{C}\right)={T},\:{i}.{e}\:\left({p}\wedge{q}\right)\Rightarrow{r}\:\:{is}\:{a}\:{tautology}. \\ $$$${Without}\:{truth}\:{tables},\:{I}'{ll}\:{attempt}\:{to}\:{use}\:{the}\:{laws}\:{of}\:{Boolean}\:{algebra} \\ $$$${to}\:{simplify}\:{the}\:{logic}\:{expression}\:{to}\:{hopefully}\:{end}\:{up}\:{having}\:{a}\:{tautology}. \\ $$$${Note}\:{that}\:\mu\Rightarrow{z}\equiv\backsim\mu\vee{z}. \\ $$$$\left[\left({A}\Rightarrow{B}\right)\wedge\left({B}\Rightarrow{C}\right)\right]\Rightarrow\left({A}\Rightarrow{C}\right)\equiv\left[\left(\backsim{A}\vee{B}\right)\wedge\left(\backsim{B}\vee{C}\right)\right]\Rightarrow\left(\backsim{A}\vee{C}\right) \\ $$$${lhs}\equiv\backsim\left[\left(\backsim{A}\vee{B}\right)\wedge\left(\backsim{B}\vee{C}\right)\right]\vee\left(\backsim{A}\vee{C}\right) \\ $$$${lhs}\equiv\left[\backsim\left(\backsim{A}\vee{B}\right)\vee\backsim\left(\backsim{B}\vee{C}\right)\right]\vee\left(\backsim{A}\vee{C}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{De}\:{Morgan}'{s}\:{law} \\ $$$${lhs}\equiv\left[\left({A}\wedge\backsim{B}\right)\vee\left({B}\wedge\backsim{C}\right)\right]\vee\left(\backsim{A}\vee{C}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{De}\:{Morgan}'{s}\:{law},\:{Double}\:{negation}\:{law} \\ $$$${lhs}\equiv\left[\backsim{A}\vee\left({A}\wedge\backsim{B}\right)\right]\vee\left[{C}\vee\left({B}\wedge\backsim{C}\right)\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Commutative}\:{law},{Associative}\:{law} \\ $$$${lhs}\equiv\left[\left(\backsim{A}\vee{A}\right)\wedge\left(\backsim{A}\vee\backsim{B}\right)\right]\vee\left[\left({C}\vee{B}\right)\wedge\left({C}\vee\backsim{C}\right)\right]\:\:\:\:{Distribution}\:{law} \\ $$$${lhs}\equiv\left[{T}\wedge\left(\backsim{A}\vee\backsim{B}\right)\right]\vee\left[\left({C}\vee{B}\right)\wedge{T}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Complementary}\:{law} \\ $$$${lhs}\equiv\left(\backsim{A}\vee\backsim{B}\right)\vee\left({C}\vee{B}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Identity}\:{law} \\ $$$${lhs}\equiv\left(\backsim{B}\vee{B}\right)\vee\left({C}\vee\backsim{A}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Commutative}\:{law},{Associative}\:{law} \\ $$$${lhs}\equiv{T}\vee\left({C}\vee\backsim{A}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Complementary}\:{law} \\ $$$${lhs}\equiv{T}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Domination}\:{law}\: \\ $$$${The}\:{simplification}\:{of}\:{the}\:{Boolean}\:{statement}\:{has}\:{given}\:{us}\:{a}\:{constant} \\ $$$${truth}\:{value}\:{of}\:{T},\:{so}\:{the}\:{statement}\:{is}\:{perpetually}\:{true}.\:{Hence},{a}\:{transitive} \\ $$$${property}\:{of}\:{the}\:{implication}\:{logic}\:{as}\:{been}\:{proven}.\:\blacksquare \\ $$$$ \\ $$

Commented by Yozzy last updated on 30/Aug/15

A∣B∣C∣A⇒B∣B⇒C∣A⇒C∣(A⇒B)∧(B⇒C)∣[(A⇒B)∧(B⇒C)]⇒(A⇒C) 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 The last column to the right indicates that the statement [(A⇒B)∧(B⇒C)]⇒(A⇒C) is always true. Therefore, this proves the existance of a transitive property of the implication logic. (I′ll check over my Boolean simplification.Checked.)

$${A}\mid{B}\mid{C}\mid{A}\Rightarrow{B}\mid{B}\Rightarrow{C}\mid{A}\Rightarrow{C}\mid\left({A}\Rightarrow{B}\right)\wedge\left({B}\Rightarrow{C}\right)\mid\left[\left({A}\Rightarrow{B}\right)\wedge\left({B}\Rightarrow{C}\right)\right]\Rightarrow\left({A}\Rightarrow{C}\right) \\ $$$$\mathrm{0}\:\:\:\:\mathrm{0}\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{0}\:\:\:\:\mathrm{0}\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{0}\:\:\:\:\mathrm{1}\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{0}\:\:\:\:\mathrm{1}\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\:\:\:\:\mathrm{0}\:\:\mathrm{0}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\:\:\:\:\mathrm{0}\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\:\:\:\:\mathrm{1}\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\:\:\:\:\mathrm{1}\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$${The}\:{last}\:{column}\:{to}\:{the}\:{right}\:{indicates}\:{that}\:{the}\:{statement} \\ $$$$\left[\left({A}\Rightarrow{B}\right)\wedge\left({B}\Rightarrow{C}\right)\right]\Rightarrow\left({A}\Rightarrow{C}\right)\:{is}\:{always}\:{true}.\:{Therefore},\:{this}\:{proves} \\ $$$${the}\:{existance}\:{of}\:{a}\:{transitive}\:{property}\:{of}\:{the}\:{implication}\:{logic}. \\ $$$$\left({I}'{ll}\:{check}\:{over}\:{my}\:{Boolean}\:{simplification}.{Checked}.\right) \\ $$

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

THANKSSsssssss! Yozzian=^(?) Yozzy

$$\boldsymbol{\mathrm{THANKS}}\mathrm{Ss}{ssssss}! \\ $$$$\mathrm{Yozzian}\overset{?} {=}\mathrm{Yozzy} \\ $$

Commented by Yozzy last updated on 30/Aug/15

I use my cell phones (2) and my tablet. I′ve been the only one commenting on this question. I used my tablet to answer the question because of the long truth table.

$${I}\:{use}\:{my}\:{cell}\:{phones}\:\left(\mathrm{2}\right)\:{and}\:{my}\:{tablet}.\:{I}'{ve}\:{been}\:{the}\:{only}\:{one}\:{commenting} \\ $$$${on}\:{this}\:{question}.\:{I}\:{used}\:{my}\:{tablet}\:{to}\:{answer}\:{the}\:{question}\:{because}\:{of}\:{the} \\ $$$${long}\:{truth}\:{table}. \\ $$

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

I have also more than one device and I use two ID′s: Rasheed Soomro and Rasheed Ahmad

$${I}\:{have}\:{also}\:{more}\:{than}\:{one}\:{device}\:{and}\:{I}\:{use}\:{two}\:{ID}'{s}: \\ $$$$\mathrm{Rasheed}\:\mathrm{Soomro}\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\mathrm{Rasheed}\:\mathrm{Ahmad} \\ $$

Commented by 112358 last updated on 30/Aug/15

I see.

$${I}\:{see}. \\ $$

Commented by 123456 last updated on 30/Aug/15

XD 2^n_i

$$\mathrm{XD}\:\mathrm{2}^{{n}_{{i}} } \\ $$

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

XD 2^n_i ???

$$\mathrm{XD}\:\mathrm{2}^{{n}_{{i}} } \:??? \\ $$

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