Question

I need help on the blue highlighted questions and 20 from the last picture. Our professor doesn’t want a truth table. He wants a proof.

In Exercises 13-24, use propositional logic to prove that the argument is valid. 13. (A VB')' A(BC) → (A' AC) 14. A' A( B A) →B' 15. (A →B) A [A → ( B C)] → ( AC ) 16.[( CD) →→[( CD) →D] 17. A' (A VB) →B Section 1.2 Propositional Logic 18. (A → ( B C)] A (AV D') AB ( D C ) 19. (A' B') ABA ( A C ) 20. (A →B) A[B ( CD)] A[A ( BC)] → ( 4D) 21. [A ( B C)] → [B (A →C)] 22. (A AB) → (A →B')' 23. ( AC) A( CB') ABA' 24. [A → (BVC)] AC' → (A → B)

Answer 1

20) (A $\rightarrow$ B) $\wedge$ [B $\rightarrow$(C $\rightarrow$D)] $\wedge$ [A $\rightarrow$(B $\rightarrow$C)] $\rightarrow$(A $\rightarrow$D)

$\equiv$ (¬A V B) $\wedge$ [¬B V (C $\rightarrow$D)] $\wedge$ ¬[¬A V (B $\rightarrow$C)] V (A $\rightarrow$D)  { Law of Implies (P$\rightarrow$Q) = ¬PVQ }

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ ¬[¬A V (¬B VC)] V (¬A V D)  { Law of Implies (P$\rightarrow$Q) = ¬PVQ }

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [¬(¬A) $\wedge$ ¬(¬B VC)] V (¬A V D)  {By De Morgan's law ¬ (PVQ) = ¬P ¬Q}

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [A $\wedge$ (¬(¬B) $\wedge$¬C)] V (¬A V D)   {By De Morgan's law ¬ (¬A) = A}

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [A $\wedge$ (B $\wedge$¬C)] V (¬A V D)    {By De Morgan's law ¬ (¬B) = B}

$\equiv$ [¬B V (¬C VD)] $\wedge$ (¬A V B) $\wedge$ [(B $\wedge$¬C) $\wedge$​​​​​​​ A ] V (¬A V D) { Commutative law A B = B A }

$\equiv$ [¬B V (¬C VD)] $\wedge$ ¬A V (B​​​​​​​ $\wedge$ B) $\wedge$​​​​​​​¬C $\wedge$​​​​​​​ (A V ¬A) V D {Associative law (A B) C = A (B C)}

$\equiv$ [¬B V (D V ¬C)] $\wedge$ ¬A V (B​​​​​​​) $\wedge$​​​​​​​¬C $\wedge$​​​​​​​ (T) V D  { we know that A V A = A }

$\equiv$ [¬B V (D V ¬C)] $\wedge$ ¬C $\wedge$​​​​​​​ ¬A V (B​​​​​​​) $\wedge$​​​​​​​ (T) V D  { Commutative law A V B = B V A }

$\equiv$ ¬B V D V (¬C $\wedge$ ¬C) $\wedge$​​​​​​​ (B​​​​​​​) V ¬A $\wedge$​​​​​​​ (T) V D {Associative law (A B) C = A (B C)}

$\equiv$ ¬B V D V (¬C) $\wedge$​​​​​​​ (B​​​​​​​) V ¬A $\wedge$​​​​​​​ (T) V D { we know that ¬A $\wedge$ ¬A = ¬A }

$\equiv$ D V (¬C) V ¬B  $\wedge$​​​​​​​ (B​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D

$\equiv$ D V (¬C) V (¬B  $\wedge$​​​​​​​ B​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D  {Associative law (A V B) V C = A V (B V C)}

$\equiv$ D V (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D {We know that ¬A F = F}

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D V D { Commutative law A V B = B V A }

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V (D V D)  

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V (D)  {Associative law (A V A) = A }

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T V D)

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T)  {We know that A V T = T}

$\equiv$ (¬C) V (T) $\wedge$ (F​​​​​​​) V (¬A) ​​​​​​​ { Commutative law A V B = B V A }

$\equiv$ (¬C V T) $\wedge$​​​​​​​ (¬A) V​​​​​​ (F​​​​​​​)  { Commutative law A V B = B V A }

$\equiv$ (T) $\wedge$ (¬A) V​​​​​​ (F​​​​​​​)  {We know that ¬A V T = T}

$\equiv$ (¬A) $\wedge$ (T) V (F) ​​​​​​​ ​​​​​​​

$\equiv$ (¬A) $\wedge$ (F) V (T) ​​​​​​​ ​​​​​​​{ Commutative law A V B = B V A }

$\equiv$ (¬A $\wedge$ F) V (T) ​​​​​​​ ​​​​​​​ {Associative law (A V B) V C = A V (B V C)}

$\equiv$ (F) V (T) ​​​​​​​ ​​​​​​​ {We know that ¬A F = F}

$\equiv$ T

The given Argument is Valid

Answer 2

20) (A $\rightarrow$ B) $\wedge$ [B $\rightarrow$(C $\rightarrow$D)] $\wedge$ [A $\rightarrow$(B $\rightarrow$C)] $\rightarrow$(A $\rightarrow$D)

$\equiv$ (¬A V B) $\wedge$ [¬B V (C $\rightarrow$D)] $\wedge$ ¬[¬A V (B $\rightarrow$C)] V (A $\rightarrow$D)  { Law of Implies (P$\rightarrow$Q) = ¬PVQ }

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ ¬[¬A V (¬B VC)] V (¬A V D)  { Law of Implies (P$\rightarrow$Q) = ¬PVQ }

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [¬(¬A) $\wedge$ ¬(¬B VC)] V (¬A V D)  {By De Morgan's law ¬ (PVQ) = ¬P ¬Q}

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [A $\wedge$ (¬(¬B) $\wedge$¬C)] V (¬A V D)   {By De Morgan's law ¬ (¬A) = A}

$\equiv$ (¬A V B) $\wedge$ [¬B V (¬C VD)] $\wedge$ [A $\wedge$ (B $\wedge$¬C)] V (¬A V D)    {By De Morgan's law ¬ (¬B) = B}

$\equiv$ [¬B V (¬C VD)] $\wedge$ (¬A V B) $\wedge$ [(B $\wedge$¬C) $\wedge$​​​​​​​ A ] V (¬A V D) { Commutative law A B = B A }

$\equiv$ [¬B V (¬C VD)] $\wedge$ ¬A V (B​​​​​​​ $\wedge$ B) $\wedge$​​​​​​​¬C $\wedge$​​​​​​​ (A V ¬A) V D {Associative law (A B) C = A (B C)}

$\equiv$ [¬B V (D V ¬C)] $\wedge$ ¬A V (B​​​​​​​) $\wedge$​​​​​​​¬C $\wedge$​​​​​​​ (T) V D  { we know that A V A = A }

$\equiv$ [¬B V (D V ¬C)] $\wedge$ ¬C $\wedge$​​​​​​​ ¬A V (B​​​​​​​) $\wedge$​​​​​​​ (T) V D  { Commutative law A V B = B V A }

$\equiv$ ¬B V D V (¬C $\wedge$ ¬C) $\wedge$​​​​​​​ (B​​​​​​​) V ¬A $\wedge$​​​​​​​ (T) V D {Associative law (A B) C = A (B C)}

$\equiv$ ¬B V D V (¬C) $\wedge$​​​​​​​ (B​​​​​​​) V ¬A $\wedge$​​​​​​​ (T) V D { we know that ¬A $\wedge$ ¬A = ¬A }

$\equiv$ D V (¬C) V ¬B  $\wedge$​​​​​​​ (B​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D

$\equiv$ D V (¬C) V (¬B  $\wedge$​​​​​​​ B​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D  {Associative law (A V B) V C = A V (B V C)}

$\equiv$ D V (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D {We know that ¬A F = F}

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V D V D { Commutative law A V B = B V A }

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V (D V D)  

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T) V (D)  {Associative law (A V A) = A }

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T V D)

$\equiv$ (¬C) V (F​​​​​​​) V (¬A) $\wedge$​​​​​​​ (T)  {We know that A V T = T}

$\equiv$ (¬C) V (T) $\wedge$ (F​​​​​​​) V (¬A) ​​​​​​​ { Commutative law A V B = B V A }

$\equiv$ (¬C V T) $\wedge$​​​​​​​ (¬A) V​​​​​​ (F​​​​​​​)  { Commutative law A V B = B V A }

$\equiv$ (T) $\wedge$ (¬A) V​​​​​​ (F​​​​​​​)  {We know that ¬A V T = T}

$\equiv$ (¬A) $\wedge$ (T) V (F) ​​​​​​​ ​​​​​​​

$\equiv$ (¬A) $\wedge$ (F) V (T) ​​​​​​​ ​​​​​​​{ Commutative law A V B = B V A }

$\equiv$ (¬A $\wedge$ F) V (T) ​​​​​​​ ​​​​​​​ {Associative law (A V B) V C = A V (B V C)}

$\equiv$ (F) V (T) ​​​​​​​ ​​​​​​​ {We know that ¬A F = F}

$\equiv$ T

The given Argument is Valid