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Given the following arguments:

$ \tag A (R \to \neg S) \land (T \to \neg U)$

$ \tag B (V\to \neg W) \land (X \to \neg Y)$

$ \tag C (T \to W) \land (U \to S)$

$ \tag D V \lor R $

$ \therefore \neg T \lor \neg U $

how to prove this using the standard rules of inference?

2 Answers 2

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One expects this to follow more or less directly from $(A)$ because $p\to q\iff \neg p \lor q$. However, there is no tertium no datur. One has to find ones luch starting from (D) by case analysis:

By simplification from (A): $\tag{1}R\to\neg S$ and $\tag2 T\to\neg U$ By simplification from (C): $\tag{3}T\to W$ and $\tag4 U\to S$ By modus ponens from (1): $\tag 5R\vdash \neg S$ By modus tollens from (4) and (5): $\tag 6 R\vdash \neg U$ By addition from (6) $\tag 7 R\vdash \neg T \lor \neg U$ By decuction theorem from (6): $\tag 8 R\to (\neg T \lor \neg U)$ By simplification from (B): $\tag 9 V\to\neg W$ By modus ponens from (9): $\tag{10} V\vdash \neg W$ By modus tollens from (10) and (3): $\tag{11}V\vdash \neg T$ By addition from (11): $\tag{12}V\vdash \neg T\lor\neg U$ By deduction theorem from (12) $\tag{13}V\to( \neg T\lor\neg U)$ By case analysis from (D), (13) and (8): $\neg T \lor \neg U_\blacksquare$

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    Thanks, but I am not accustomed with sequent notation, so I couldn't understand some parts of your solution. Could you please edit it or perhaps suggest me some reference?2012-10-25