Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Set Theory, Logic, Probability, Statistics
Proving Similar Statements Using Proof Alteration
Reply to thread
Message
[QUOTE="mXSCNT, post: 2201312, member: 164922"] Suppose you have a formal proof of a statement S, but you want to prove a statement S' which is in some way similar to S. The question is, how might a computer slightly alter the proof of S in such a way that it becomes a proof of S'? Here's an example (using v for disjunction, n for conjunction). Skip it if you find it tedious. [code] 1. (A v B) n C (assume.) | 2. (A v B) (1, conjunction elimination) | 3. (A v B) v (A v B) (2, disjunction introduction) | 4. C (1, conjunction elimination) | 5. ((A v B) v (A v B)) n C (3,4, conjunction introduction) 6. ((A v B) n C) -> (((A v B) v (A v B)) n C) (1,5,conditional proof) [/code] Now suppose that instead of S = ((A v B) n C) -> (((A v B) v (A v B)) n C), we wanted to prove S' = (~(A n B) n C) -> (((~A v ~B) v D) n C). This could be done as follows: [code] 1. ~(A n B) n C (assume.) | 2. ~(A n B) (1, conjunction elimination) | 3. ~A v ~B (2, DeMorgan) | 4. (~A v ~B) v D (3, disjunction introduction) | 5. C (1, conjunction elimination) | 6. ((~A v ~B) v D) n C (4,5, conjunction introduction) 7. (~(A n B) n C) -> (((~A v ~B) v D) n C) (1,6,conditional proof) [/code] The proof for S' has been derived from the proof for S, by replacing the premise (A v B) n C with the premise ~(A n B) n C, applying DeMorgan's rule after step 2, and introducing D instead of (A v B) in the disjunction introduction. [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Set Theory, Logic, Probability, Statistics
Proving Similar Statements Using Proof Alteration
Back
Top