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(I^#h/a,u"rKqgF0aL4k/k3h*hepN5fHpWYBR80&F0g1#adpo4Dj15#.VB& They read "p if and only if q" and are denoted \(p \leftrightarrow q\) or "p iff q", which is logically equivalent to \((p \to q) \wedge (q \to p)\). @beq06Cl=Eb/j$Eb-A4F`[t$F`8H\1*A5hEb/`lA7]h#E /Flags 4 5.423 0 Td (1) Tj 0.987 0 Td (called) Tj /Filter /ASCII85Decode -c*De"tE/$X>bAt_j]52'EmJV[+-`g!T\PIjS+g4&$t!^6:m;^]!3Gshe=Q3`:p)Otq3Yq@`7Xk\=N[s1.oRFh"3e"T[8L_$,P:X[-l:Un?Ht72]GHGho^lMCXU ", If a person is looking for a house with 4 bedrooms or a short commute, a real estate agent might present houses with either 4 bedrooms or a short commute or both 4 bedrooms and a short commute. /F1 1 Tf e) p p f) (p q) (p q). 1.985 0 Td (osition) Tj 9u+>Gc2+>6DsD.7$a+E2IF$=n9u+>Gc4+>6Q*+E2IF$=n9u+>Gc5+>6Z-B5M'"E-670A9Di60fD!60 Q q /Type /Font Let's consider a tautology first, and then a contradiction: Consider the statement "\((2 = 3) \vee (2 \ne 3)\)": Let's make a truth table for general case \(p \vee (\neg p)\): As you can see, no matter what we do, this statement is always true. :7&:J!lU($\a*M;= 0.843 0 Td (and) Tj BT +j/I[FTEp;^49ca35!WhAmgk`hO=paPhUgJqZ2+#hd5+M+mO2XSJG0>osWJ)@[YH[*\DA;.F/M%Q)08:5CJJ-,fIZBu7/45F%$@RM#V:i0\Oor#c"h>?M,"4E+VdS? 1.095 0 Td (a) Tj "c)0;p+i"7V#rJLB:.>jVlclII/\/)Q0nA_Q^e*[#Kolj,Ut0VPrFLnE[,fg$p;- Tautology: A statement that is always true, and a truth table yields only true results. 39 0 obj Q q &iQ:c*EftB`L*Pgnuf%$>uE$G*VU/Wnf;9dc;1r4;kWnA%"&/OMT9E>R2W1R1F*/g^T/Zs*fee2lsV Pigs can fly. 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