Symbolic Logic
Symbolic Logic
Symbolic Logic works on the basis of 19 main rules and a violation of the rules results in Fallacy:
The Rules are:
Sr.No | Rule | Symbolization | Reading of the Symbolized Rule |
1 | Modus Ponen | p⊃q, p, ∕q | p implies q
p Therefore, q |
2 | Modus Tonen | p⊃q, ~q, ∕~p
|
p implies q
not q Therefore, not p |
3 | Hypothetical Syllogism | p=q, q=r, /p=r
|
p is equal to q
q is equal to r Therefore, p is equal to r
|
4 | Disjunctive Syllogism | ~p ∨ q, ~p/ q | Not p or q
Not p Therefore, q |
5 | Constructive Dilemma | (p⊃q). (r⊃s), p ∨ r/ q ∨ s
|
(p implies q) and (r implies s)
p or r Therefore, q or s |
6 | Destructive Dilemma | (p⊃q).(r⊃s), ~q ∨ ~s/ ~p ∨ ~r
|
( p implies q) and ( r implies s)
Not q or not s Therefore, not p or not r |
7 | Simplification | p.q/ p | p and q
Therefore, p |
8 | Conjunction | p, q/ p.q | p
q Therefore, p and q |
9 | Addition | p/ p ∨q
|
P
Therefore, p or q |
10 | DeMorgan`s Law
|
1. ~(p.q)= (~pv~q)
2. ~(pvq)=( ~p. ~q) |
1. Negation p and q is equivalent to (negation p or negation q)
2. Negation p or q is equivalent to (negation p and negation q) |
11 | Commutation
|
1. (p,q)=(q.p)
2. (pvq)=(qvp) |
1. p and q is equivalent to q and p
2. p or q is equivalent to q or p |
12 | Association
|
1. [(p.q).r]=[p.(q.r)]
2. [(pvq)vr]=[pv(qvr)] |
1. (p and q) and r are equivalent to p and (q and r)
2. (p or q) and r are equivalent to p and (q or r)
|
13 | Distribution
|
1. [p.(qvr)]=[(p.q)v(p.r)]
2. [pv(q.r)]=[(pvq).(pvr)] |
1.[ p and (q or r)] is equivalent to [(p and q) or (p and r)]
2. [p or (q and r)] is equivalent to [(p or q) and (p or r)] |
14 | Double Negation
|
~~p=p | Not Not p is equivalent p |
15 | Transposition
|
(p⊃q)= (~q⊃~p) | (p implies q) is equivalent to (not q implies not p) |
16 | Material Implication
|
(p⊃q)=(~pvq) | (p implies q) is equivalent to (not p or q) |
17 | Material Equivalence
|
1. (p=q)=[(p⊃q). (q⊃p)]
2. (p=q)=[(p.q) v (~p. ~q)] |
1. (p equals q) is equivalent to [(p implies q) and (q implies p)]
2. (p equals q) is equivalent to [ (p and q) or (not p and not q)
|
18 | Exportation
|
[(p.q) ⊃r]=[p⊃(q⊃r)] | [(p and q) implies r] is equivalent to [p implies ( q implies r)] |
19 | Tautology
|
1. p= (p.p)
2. p=(pvp) |
1. p is equivalent to (p and p)
2. p is equivalent to (p or p) |