Exercise logic.propositional.dnf.unicode
Description
Proposition to DNF (unicode support)
Firsts
Rule | logic.propositional.defimpl |
Location | [0,0,0,0] |
Term | "Nothing" |
Focus | "Nothing" |
Environment | |
![](http://ideas.cs.uu.nl/images/external.png)
((¬(¬q ∨ p) ↔ ((r ↔ s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))) ∧ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))
ready: no
Rule | logic.propositional.defequiv |
Location | [0,0,1,0] |
Term | "Nothing" |
Focus | "Nothing" |
Environment | |
![](http://ideas.cs.uu.nl/images/external.png)
((¬(q → p) ↔ ((r ∧ s) ∨ (¬r ∧ ¬s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))) ∧ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))
ready: no
Rule | logic.propositional.defequiv |
Location | [0,0] |
Term | "Nothing" |
Focus | "Nothing" |
Environment | |
![](http://ideas.cs.uu.nl/images/external.png)
((¬(q → p) ∧ ((r ↔ s) ∨ ¬s)) ∨ (¬¬(q → p) ∧ ¬((r ↔ s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s)) ∨ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))) ∧ (¬(q → p) ↔ ((r ↔ s) ∨ ¬s))
ready: no