1. Determine whether this proposition is a tautology.
[(p → q) ∧ (q → r)] → (p→ r)
[(p→q)∧(q→r)]→(p→r)=¬[(¬p∨q)∧(¬q∨r)]∨(¬p∨r)==¬(¬p∨q)∨¬(¬q∨r)∨¬p∨r==(p∧¬q)∨(q∧¬r)∨¬p∨r==¬q∨¬p∨q∨r=T\left[ \left( p\rightarrow q \right) \land \left( q\rightarrow r \right) \right] \rightarrow \left( p\rightarrow r \right) =\lnot \left[ \left( \lnot p\lor q \right) \land \left( \lnot q\lor r \right) \right] \lor \left( \lnot p\lor r \right) =\\=\lnot \left( \lnot p\lor q \right) \lor \lnot \left( \lnot q\lor r \right) \lor \lnot p\lor r=\\=\left( p\land \lnot q \right) \lor \left( q\land \lnot r \right) \lor \lnot p\lor r=\\=\lnot q\lor \lnot p\lor q\lor r=T[(p→q)∧(q→r)]→(p→r)=¬[(¬p∨q)∧(¬q∨r)]∨(¬p∨r)==¬(¬p∨q)∨¬(¬q∨r)∨¬p∨r==(p∧¬q)∨(q∧¬r)∨¬p∨r==¬q∨¬p∨q∨r=T
The formula is a tautology
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