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Exercícios de semântica formal para a Lógica Proposicional Clássica
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Revision as of 17:07, 25 September 2020 by
Greati
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→\Gamma_1 \vdash \varphi \to \psi; \Gamma_2 \vdash \varphi / \Gamma_1,\Gamma_2 \vdash \psi
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Contents
1
Validade de sequentes
1.1
¬
(
r
↔
p
)
{\displaystyle \neg (r\leftrightarrow p)}
▶
q
→
¬
p
{\displaystyle q\to \neg p}
1.2
(
p
∧
q
)
→
r
{\displaystyle (p\land q)\to r}
▶
q
→
¬
p
{\displaystyle q\to \neg p}
1.3
¬
(
r
↔
p
)
,
(
p
∧
q
)
→
r
{\displaystyle \neg (r\leftrightarrow p),(p\land q)\to r}
▷
q
→
¬
p
{\displaystyle q\to \neg p}
2
Correção de regras
2.1
(
→
E
)
Γ
1
⊢
φ
→
ψ
;
Γ
2
⊢
φ
/
Γ
1
,
Γ
2
⊢
ψ
{\displaystyle (\to E)\Gamma _{1}\vdash \varphi \to \psi ;\Gamma _{2}\vdash \varphi \quad /\quad \Gamma _{1},\Gamma _{2}\vdash \psi }
2.2
(
→
I
)
Γ
1
,
φ
⊢
ψ
/
Γ
⊢
φ
→
ψ
{\displaystyle (\to I)\Gamma _{1},\varphi \vdash \psi \quad /\quad \Gamma \vdash \varphi \to \psi }
3
Para reflexão
4
Veja também
5
Links externos
Validade de sequentes
¬
(
r
↔
p
)
{\displaystyle \neg (r\leftrightarrow p)}
▶
q
→
¬
p
{\displaystyle q\to \neg p}
(
p
∧
q
)
→
r
{\displaystyle (p\land q)\to r}
▶
q
→
¬
p
{\displaystyle q\to \neg p}
¬
(
r
↔
p
)
,
(
p
∧
q
)
→
r
{\displaystyle \neg (r\leftrightarrow p),(p\land q)\to r}
▷
q
→
¬
p
{\displaystyle q\to \neg p}
Correção de regras
(
→
E
)
Γ
1
⊢
φ
→
ψ
;
Γ
2
⊢
φ
/
Γ
1
,
Γ
2
⊢
ψ
{\displaystyle (\to E)\Gamma _{1}\vdash \varphi \to \psi ;\Gamma _{2}\vdash \varphi \quad /\quad \Gamma _{1},\Gamma _{2}\vdash \psi }
(
→
I
)
Γ
1
,
φ
⊢
ψ
/
Γ
⊢
φ
→
ψ
{\displaystyle (\to I)\Gamma _{1},\varphi \vdash \psi \quad /\quad \Gamma \vdash \varphi \to \psi }
Para reflexão
Veja também
Links externos