Determine whether each of these compoundpropositions is satisfiable. a) (p∨q∨¬r)∧(p∨¬q∨¬s)∧(p∨¬r∨¬s)∧ (¬p ∨¬q∨¬s)∧(p∨q∨¬s) b) (¬p∨¬q∨r)∧(¬p∨q∨¬s)∧(p∨¬q∨¬s)∧ (¬p ∨¬r ∨¬s)∧(p∨q∨¬r)∧(p∨¬r∨¬s) c) (p∨q∨r)∧(p∨¬q∨¬s)∧(q∨¬r∨s) ∧(¬p∨ r ∨s) ∧(¬p∨q∨¬s)∧(p∨¬q∨¬r) ∧(¬p∨ ¬q ∨s)∧(¬p∨¬r∨¬s)

See below

Step-by-step explanation:

A compound proposition is satisfiable if there exists a combination of truth values of the component propositions that make the compound proposition true.

A compound conjunction p1∧p2∧p3∧...∧pn is true iff pi is true for all i.

A compound disjunction p1∨p2∨p3∨...∨pn is true iff pi is true for some i.

a) (p∨q∨¬r)∧(p∨¬q∨¬s)∧(p∨¬r∨¬s)∧(¬p∨¬q∨¬s)∧(p∨q∨¬s) is satisfiable.

Assume that p is true, q is true, r is false and s is false.

Since p is true, then (p∨q∨¬r), (p∨¬q∨¬s) and (p∨¬r∨¬s) are true.

Now, s is false, then ¬s is true. Then (¬p∨¬q∨¬s) and (p∨q∨¬s) are true.

Combine both statements to obtain that  (p∨q∨¬r)∧(p∨¬q∨¬s)∧(p∨¬r∨¬s)∧(¬p∨¬q∨¬s)∧(p∨q∨¬s) is true.

b) (¬p∨¬q∨r)∧(¬p∨q∨¬s)∧(p∨¬q∨¬s)∧(¬p∨¬r∨¬s)∧(p∨q∨¬r)∧(p∨¬r∨¬s) is satisfiable

Assume that p is false, q is false, r is false and s is false.

Since p is false, ¬p is true, then (¬p∨¬q∨r), (¬p∨q∨¬s) and (¬p∨¬r∨¬s) are true.

Since s is false, ¬s is true, then (p∨¬q∨¬s) and (p∨¬r∨¬s) are true.

Since s is false, ¬s is true, then (p∨q∨¬r) is true.

Combining all of these, the compound proposition is satisfiable (each component of the disjunction is true).

c) (p∨q∨r)∧(p∨¬q∨¬s)∧(q∨¬r∨s)∧(¬p∨r∨s)∧(¬p∨q∨¬s)∧(p∨¬q∨¬r)∧(¬p∨¬q ∨s)∧(¬p∨¬r∨¬s) is satisfiable.

Assume that p is true, q is true, s is true and r is false.

Since p is true, (p∨q∨r), (p∨¬q∨¬s) and (p∨¬q∨¬r) are true.

Since s is true, (q∨¬r∨s), (¬p∨r∨s) and (¬p∨¬q ∨s) are true.

Since ¬r is true, (p∨¬q∨¬r) and  (¬p∨¬r∨¬s) are true.

Since q is true, (¬p∨q∨¬s) is true.

Combining all of the above, the compound proposition is satisfiable.

Remark: there are unsatisfiable propositions, but proving that a proposition is unsatisfiable requires a general argument. For example, (p∧q)∧(¬p) is unsatisfiable, since it contains the contradiction (p∧¬p).

da freeck kinda math is this just drop out

step-by-step explanation:

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.