Determine whether this proposition is a tautology.
[(p → q) Ù (q → r)] → (p→ r)
Let p=1,q=0,r=0. Then
p→q=0p→q=0q→r=1(p→q)∨(q→r)=1((p→q)∨(q→r))→(p→r)=0p\rightarrow q=0\\ p\rightarrow q=0\\ q\rightarrow r=1\\ (p\rightarrow q)\lor (q\rightarrow r)=1\\ ((p\rightarrow q)\lor (q\rightarrow r))\rightarrow (p\rightarrow r)=0p→q=0p→q=0q→r=1(p→q)∨(q→r)=1((p→q)∨(q→r))→(p→r)=0
The formula is not a tautology.
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