SOLUTION: 1. A→B 2. A→~(C∨B) ∴ ~A 1. (A&~B)&(~C∨D) 2. ~((D&A)&~B) ∴ ~C

Algebra ->  Proofs -> SOLUTION: 1. A→B 2. A→~(C∨B) ∴ ~A 1. (A&~B)&(~C∨D) 2. ~((D&A)&~B) ∴ ~C       Log On


   

Question 935456: 1. A→B
2. A→~(C∨B) ∴ ~A
1. (A&~B)&(~C∨D)
2. ~((D&A)&~B) ∴ ~C
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1.	A→B
2.	A→~(C∨B) ∴ ~A 

3.      A→(~C&~B)      2, DeMorgan's theorem
4.      A→~B           3, Decomposition (simplification) of a conjunction
5.      ~B→~A          1, Modus Tollens
6.      ~~B→~A         4, Modus Tollens
7.      B→~A          7, Double negation
8.      B∨~B          Tautology (law of excluded middle)
9.       ~A             7,5,8, disjunctive elimination (disjunctive syllogism)      

 

1.	(A&~B)&(~C∨D)
2.	~[(D&A)&~B] ∴ ~C

3.      ~[D&(A&~B)]      2,Association
4.      ~D∨~(A&~B)        3,DeMorgan's theorem 
5.      A&~B             1,Decomposition (simplification) of a conjunction  
6.      ~D               4,5, disjunctive elimination (disjunctive syllogism) 
7.      ~C∨D             1,Decomposition (simplification) of a conjunction
8.      ~C               7,6, disjunctive elimination (disjunctive syllogism) 

Edwin