MATH 140A Lecture 2: Note Sep 28, 2018
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Iii hay fy at f x y xy3 yz. F a successor of n in1n called nil. Nen is not a successor of any element in n. If n m e n have the same successorthen n m. Not enough iii 4 s"s lets be a subset of 1n suchthat. Assume that 5 contains htt whenever it contains n then. 1 p ii priti base step is true whenever pn is true. 1 3 5 7 c9 25 52 n2. Pnm 1 3 5 t gen c t h 2 bypn anti. So by p. m. i all ph"s are true 0. Proof of the p m. i i ein i pi. Applying peano"s as we conclude that 5 in then nties anti is true. 0 is true is true when pn is true instead of 1 n eno. Ex i pno ii priti n2h31i 13 t. Prop i n z n3 for n and for any mi6.
