Note: $\neg$ means 'not', $\rightarrow$ is 'conditional', $\land$ is 'and', $\lor$ means '(inclusive) or'.
Prove: $[\neg D \lor (A \land B)] \rightarrow[(J \rightarrow \neg A) \rightarrow (D \rightarrow \neg J)]$ using Reductio ad absurdum (RAA) or conditional proof (CP).
- $\neg[(J \rightarrow \neg A) \rightarrow (D \rightarrow \neg J)]$ (Assume for RAA)
- $(J \rightarrow \neg A) \land \neg(D \rightarrow \neg J)$ (Equation 1 is only false if the the left is true and the right is false)
- $\neg(D \rightarrow \neg J)$ (simplification of equation 2, I think)
- $D \land \neg\neg J$ (the only way for eq 3 to be true is if $D$ is true and $\neg J$ is false ).
- $D$ (simplification of 4)
- $\neg\neg J$ (simplification of 4)
- $J$ (double negative of 6).
- $J \rightarrow \neg A$ (simplification of left term equation 2
- $\neg A$ (eq 7, 8, $J$ is true, so $\neg A$ is true).
- $\neg(A \land B)$ (if $A$ is false, $A$ and anything is false, negate that to get a true)
- $\neg[\neg D \vee (A \land B)]$ ($A\land B$ is false, from eq 10, and $D$ is true from eq 5, $\neg D$ is false, false and false is false)
12 $\neg [(J \rightarrow \neg A) \rightarrow (D \rightarrow \neg J)] \rightarrow \neg[\neg D \lor (A \land B)]$ (1-11, RAA)
13 $[\neg D \lor (A \land B)] \rightarrow [(J \rightarrow \neg A) \rightarrow (D \rightarrow \neg J)]$ (contrapositive of eq 12)
What's wrong with the above proof? The teacher said it did not use RAA or CP and gave 0 points. Why not?