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This sequent is provable.
\begin{logicproof}{2} (s \lor \lnot u) \imp t & \prem \\ s \lor \lnot s & $\LEM$ \\ \begin{subproof} s & \assum \\ s\lor \lnot u & $\ori{1} 3$ \\ t & $\impe 4, 1$ \\ (\lnot s \land u) \lor t & $\ori{2} 5$ \end{subproof} \begin{subproof} \lnot s & \assum \\ u \lor \lnot u & $\LEM$ \\ \begin{subproof} u & \assum \\ \lnot s \land u & $\andi 7,9$ \\ (\lnot s\land u) \lor t & $\ori{1} 10$ \end{subproof} \begin{subproof} \lnot u & \assum \\ s \lor \lnot u & $\ori{2} 12$ \\ t & $\impe 13 1$ \\ (\lnot s\land u) \lor t & $\ori{2} 14$ \end{subproof} (\lnot s\land u) \lor t & $\ore 8, 9-11, 12-15$ \end{subproof} (\lnot s\land u) \lor t & $\ore 2, 3-6, 7-16$ \end{logicproof}
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