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




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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