- #1
Klungo
- 136
- 1
I'm reading through a text's proof on proof by contradiction. But it makes inexplicable jumps and doesn't appear to use some of the things brought up.
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.
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.