https://github.com/eser/cmpe534

CMPE534 Automated Deduction Course
https://github.com/eser/cmpe534

Last synced: 13 days ago
JSON representation

CMPE534 Automated Deduction Course

• Host: GitHub
• URL: https://github.com/eser/cmpe534
• Owner: eser
• License: gpl-2.0
• Created: 2013-11-29T22:09:52.000Z (almost 11 years ago)
• Default Branch: master
• Last Pushed: 2014-01-14T05:36:18.000Z (almost 11 years ago)
• Last Synced: 2024-05-02T05:54:21.998Z (7 months ago)
• Language: C#
• Size: 1.39 MB
• Stars: 1
• Watchers: 1
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md
  • License: LICENSE

Awesome Lists containing this project

README

        
CMPE534 Automated Deduction

===========================

This repository contains some practical implementations for CMPE534 Automated Deduction course.

Project #1: Deduction

---------------------

This project is designed to be a basic interpreter for proving theorems in propositional logic. It is basically accepts propositional statements in a specific syntax in order to prove its validity. For non-valid statements, all falsifying valuations will be extracted from counter-examples.

**To-dos:**

* 'Properties' class with 'Distribute' method needed, to resolve inside parenthesis.

* more simplifications

**Roadmap:**

* Gentzen G for First-Order Logic

* Resolution ?

**Output:**

``` bash

proposition               = (First | Second) & (A | B) & C

Dumper.Dump()             = (((First | Second) & (A | B)) & C)

Substitutor.Substitute()  = (((First | First) & (A | t)) & C)

Simplifier.Simplify()     = ((First & t) & C)

Deduction tree of: (B > C) & (D | E) -> (A & B), (C | D)

sequent = ((B > C) & (D | E)) -> (A & B), (C | D)

        sequent = (B > C), (D | E) -> C, D, A

                sequent = (D | E) -> B, C, D, A

                        sequent = D -> B, C, D, A

                                  ** axiom node **

                        sequent = E -> B, C, D, A

                                  ** counter-example node **

                sequent = C, (D | E) -> C, D, A

                          ** axiom node **

        sequent = (B > C), (D | E) -> C, D, B

                sequent = (D | E) -> B, C, D, B

                        sequent = D -> B, C, D, B

                                  ** axiom node **

                        sequent = E -> B, C, D, B

                                  ** counter-example node **

                sequent = C, (D | E) -> C, D, B

                          ** axiom node **

Formula is not valid, details below.

+ Counter-examples:

  .. E -> B, C, D, A

  .. E -> B, C, D, B

+ Falsifying valuations:

  .. Valuation #1:

  ..                 E -> t

  ..                 B -> f

  ..                 C -> f

  ..                 D -> f

  ..                 A -> f

  .. Valuation #2:

  ..                 E -> t

  ..                 B -> f

  ..                 C -> f

  ..                 D -> f

```

License

-------

See [LICENSE](LICENSE)

Contributing

------------

Contributions are welcome

[Eser Ozvataf](http://eser.ozvataf.com/)