{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture17

# lecture17 - We left off talking about how to use CP to...

This preview shows pages 1–5. Sign up to view the full content.

Uses for Conditional Proof We left off talking about how to use CP to derive biconditionals and disjunctions. To derive a biconditional p ↔ q, use conditional proof to derive p → q and q→p. Then conjoin the conditionals and use ME: p ↔ q :: (p → q) ● (q → p).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Using Conditional Proof To derive a disjunction, derive the conditional that is equivalent by ME, then use ME: p → q :: ~p v q An example: J → (K → L), J →(M →L), ~L .˙. ~J v ~(K v M)
Using Conditional Proof This derivation illustrates another case where CP can be useful: 1)  ~X v (O ● W) 2) (X O) (W X) .˙. W X 3) W Assume 4) X Assume 5) ~~X 4, DN 6) O ● W 1,5, DS 7) O 6, Simp 8) X O 4-7, CP 9) W X 2, 8, MP 10) X 3, 9, MP 11) W X 3-10, CP My initial assumption of W was not very helpful, so I used CP to  derive the antecedent of (2) and proceeded from there.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Collapsing Steps These proofs can get a little long. We can  shorten them by applying more than one rule  to a line. This is called  collapsing steps . This isn’t mandatory. If you feel more  comfortable writing out each step, I  encourage you to do so. This method is  primarily for those who have a good grasp of  derivations and want to have shorter  derivations.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}