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Proof of Propositional Logic in Discrete Mathematics
Because the words of the premise of conclusion judgment can not be simply analyzed by "propositional logic", but must rely on "predicate logic"

Define predicates:

F (x): x is a flower;

G (x): x is grass;

L(x): people who like x;

Premise:

(1) People like flowers: (ax) (f (x) → l (x)); Answer: Quantifier for full name.

(2) People don't like grass: (ax) (g (x) →┐ L (x));

Conclusion:

(3) Flowers are not grass: (ax) (f (x) →┐ g (x));

Because each judgment is aimed at all X, the influence of quantifiers can be ignored in the reasoning process. In other words, we can simply use the rules of propositional logic for reasoning. The following reasoning ignores quantifiers and variables, but it is still important to remember that without variables and quantifiers, the symbols behind them are meaningless.

Combining (1) and (2), we can get:

(F→L)∧(G→┐L)

& lt= & gt(┐F∨L)∧(┐G∨┐L)

& lt= & gt(┐f∧┐g)∩(┐f∧┐l)∩(┐g∧l)

Because:

1)┐f∧┐g = & gt; ┐F∨┐G <= & gtf→┐g;

2)┐f∧┐l = & gt; ┐f = & gt; ┐F∨┐G <= & gtf→┐g;

3)┐g∧l = & gt; ┐g = & gt; ┐F∨┐G <= & gtf→┐g;

So:

(┐f∧┐g)∨(┐f∧┐l)∨(┐g∧l)= & gt; f→┐g;

Add quantifiers and variables to get results:

( 1),(2)= & gt; (3);