Truth table calculator

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Solution

Solution provided by AtoZmath.com

Here t is used as Tautology and c is used as Contradiction

1. Prepare the truth table for Logical Expression like

1. p or q

2. p and q

3. p nand q

4. p nor q

5. p xor q

6. p => q

7. p <=> q

1. Prove the following laws using truth table

1. Commutative laws

1. p or q = q or p

2. p and q = q and p

2. Associative laws

1. (p or q) or r = p or (q or r)

2. (p and q) and r = p and (q and r)

3. De-Morgan's laws

1. not(p or q) = not p and not q

2. not(p and q) = not p or not q

4. Distributive laws

1. p and (q or r) = (p and q) or (p and r)

2. p or (q and r) = (p or q) and (p or r)

5. Negation law

1. not(not p) = p

6. Absorption laws

1. p and (p or q) = p

2. p or (p and q) = p

7. Laws for Tautology

1. p or t = t

2. p and t = t

3. p or not p = t

8. Laws for Contradiction

1. p or c = p

2. p and c = c

3. p or not p = c

9. Implication

1. p => q = (not p) or q

2. p => q = (not q) => (not p)

3. not(p => q) = p and (not q)

10. Double Implication

1. p <=> q = (p => q) and (q => p)

Example

1. Prepare the truth table for Expression : p and (q or r)=(p and q) or (p and r)

Solution:
To prove `p^^(qvvr)=(p^^q)vv(p^^r)`, we havet to first prepare the following truth table

`(1)`	`(2)`	`(3)`	`(4)=(2)vv(3)`	`(5)=(1)^^(4)`	`(6)=(1)^^(2)`	`(7)=(1)^^(3)`	`(8)=(6)vv(7)`
`p`	`q`	`r`	`qvvr`	`p^^(qvvr)`	`p^^q`	`p^^r`	`(p^^q)vv(p^^r)`
T	T	T	T `T=TvvT`	T `T=T^^T`	T `T=T^^T`	T `T=T^^T`	T `T=TvvT`
T	T	F	T `T=TvvF`	T `T=T^^T`	T `T=T^^T`	F `F=T^^F`	T `T=TvvF`
T	F	T	T `T=FvvT`	T `T=T^^T`	F `F=T^^F`	T `T=T^^T`	T `T=FvvT`
T	F	F	F `F=FvvF`	F `F=T^^F`	F `F=T^^F`	F `F=T^^F`	F `F=FvvF`
F	T	T	T `T=TvvT`	F `F=F^^T`	F `F=F^^T`	F `F=F^^T`	F `F=FvvF`
F	T	F	T `T=TvvF`	F `F=F^^T`	F `F=F^^T`	F `F=F^^F`	F `F=FvvF`
F	F	T	T `T=FvvT`	F `F=F^^T`	F `F=F^^F`	F `F=F^^T`	F `F=FvvF`
F	F	F	F `F=FvvF`	F `F=F^^F`	F `F=F^^F`	F `F=F^^F`	F `F=FvvF`

from this table, we can say that columns (5) and (8) are identical.
`:. p^^(qvvr)=(p^^q)vv(p^^r)`