- #1

Klungo

- 136

- 1

Here's the theorem and proof in the text (shortened with comment).

[tex] \mbox{Theorem: If } \Sigma \cup \{\lnot P\} \vdash \{Q\} \mbox{ and }\Sigma \cup \{\lnot P\} \vdash \{\lnot Q\} \mbox{ then } \Sigma \vdash \{P\}.[/tex]

[tex] \mbox{Proof: }\Sigma \vdash \{P \lor \lnot P\} \mbox{ ,(1) [I understand this result]. }[/tex]

[tex]\Sigma \cup \{P\} \vdash \{P\} \mbox{,(2) [By Axiom]. }[/tex]

[tex]\Sigma \vdash \{\lnot P,P\} \mbox{ ,(3) [I understand this result]. }[/tex]

[tex]\mbox{Since }\Sigma \cup \{\lnot P\} \vdash \{Q\} \mbox{ and }\Sigma \cup \{\lnot P\} \vdash \{\lnot Q\} \mbox{ ,then }\Sigma \cup \{\lnot P\} \vdash \{P\}\mbox{ ,(4) [?]. }[/tex]

[tex]\Sigma \cup \{P \lor \lnot P\} \vdash \{P\} \mbox{ ,(5) [Follows from (4)]. }[/tex]

[tex]\Sigma \vdash \{P\} \mbox{ ,(6) [Follows from (5)]. }[/tex]

It doesn't get much clearer than this in the text. There should be no errors.