論理的思考法/演習問題
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演繹推論
以下の記述は何を証明している解読すること
environ vocabularies NUMBERS, REAL_1, FINSEQ_1, VALUED_0, XBOOLE_0, NEWTON, ARYTM_3, RELAT_1, NAT_1, XXREAL_0, ARYTM_1, SUBSET_1, CARD_1, CARD_3, ORDINAL4, TARSKI, INT_2, FUNCT_1, FINSEQ_2, PRE_POLY, PBOOLE, FINSET_1, XCMPLX_0, UPROOTS, FUNCT_2, BINOP_2, SETWISEO, INT_1, FUNCOP_1, NAT_3, XREAL_0; notations TARSKI, XBOOLE_0, SUBSET_1, FINSET_1, ORDINAL1, CARD_1, NUMBERS, XCMPLX_0, XXREAL_0, XREAL_0, REAL_1, NAT_D, INT_2, RELAT_1, FUNCT_1, FUNCT_2, FINSEQ_1, FINSEQ_2, VALUED_0, PBOOLE, RVSUM_1, NEWTON, WSIERP_1, TREES_4, BINOP_2, FUNCOP_1, XXREAL_2, SETWOP_2, PRE_POLY; constructors BINOP_1, SETWISEO, NAT_D, FINSEQOP, FINSOP_1, NEWTON, WSIERP_1, BINOP_2, XXREAL_2, RELSET_1, PRE_POLY, REAL_1,CARD_1; registrations XBOOLE_0, RELAT_1, FUNCT_1, FINSET_1, NUMBERS, XCMPLX_0, XXREAL_0, NAT_1, INT_1, BINOP_2, MEMBERED, NEWTON, VALUED_0, FINSEQ_1, XXREAL_2, CARD_1, FUNCT_2, RELSET_1, ZFMISC_1, FINSEQ_2, PRE_POLY, XREAL_0, RVSUM_1; requirements NUMERALS, SUBSET, ARITHM, REAL, BOOLE; definitions TARSKI, XBOOLE_0, INT_2, NAT_D, FINSEQ_1, VALUED_0, PRE_POLY,FINSET_1,CARD_1; theorems ORDINAL1, NEWTON, NAT_1, XCMPLX_1, INT_1, CARD_4, XREAL_0, RVSUM_1, INT_2, PEPIN, FUNCT_1, CARD_2, PREPOWER, FINSEQ_1, TARSKI, XBOOLE_1, FUNCOP_1, WSIERP_1, XBOOLE_0, FINSEQ_2, FINSEQ_3, FINSEQ_4, RELAT_1, FINSOP_1, FUNCT_2, XREAL_1, XXREAL_0, NAT_D, VALUED_0, XXREAL_2, FINSET_1,PARTFUN1, PRE_POLY, CARD_1; schemes NAT_1, PRE_CIRC, FINSEQ_1, FINSEQ_2, PBOOLE, CLASSES1; begin now let Humankind be finite set, Tokyoite be Subset of Humankind, Numberofhair be Function of Tokyoite,NAT ; assume LM1: card (Tokyoite) = 12*10|^6; assume LM2: for x be object st x in Tokyoite holds Numberofhair.x <= 10|^6; LM0: 10|^6 + 1 < 12*10|^6 proof 0 < 10|^6 by PREPOWER:6; then P2: 1*10|^6 < 11* 10|^6 by XREAL_1:68; P3: 1 < 10 & 2 <= 6; then 10 < 10 |^6 by PREPOWER:13; then 1 < 10 |^6 by XXREAL_0:2,P3; then 1 < 11*10|^6 by P2,XXREAL_0:2; then P4: 1*10|^6 + 1 < 1*10|^6 + 11*10|^6 by XREAL_1:8; 1*10|^6 + 11*10|^6 = (1+11)*10|^6 ; hence thesis by P4; end; LM3: card (rng Numberofhair) <= 10|^6+1 proof now let y be object ; assume y in rng Numberofhair; then consider x be object such that A1: x in Tokyoite & y=Numberofhair.x by FUNCT_2:11; Numberofhair.x <= 10|^6 by A1,LM2; then Numberofhair.x < 10|^6+1 by NAT_1:16,XXREAL_0:2; then Numberofhair.x in Segm (10|^6+1) by NAT_1:44,A1; hence y in Segm (10|^6+1) by A1; end; then A2: rng Numberofhair c= Segm (10|^6+1) by TARSKI:def 3; then card rng Numberofhair <= card Segm (10|^6+1) by NAT_1:43; then card rng Numberofhair <= card (10|^6+1) by ORDINAL1:def 17; hence card rng Numberofhair <= (10|^6+1) ; end; LM4: card (rng (Numberofhair)) < card (Tokyoite) proof reconsider N1= card (rng (Numberofhair)) as Element of NAT ; reconsider N2= card (Tokyoite) as Element of NAT ; A1: N1<=(10|^6+1) & N2=12*10|^6 by LM1,LM3; then N1 < N2 by A1,XXREAL_0:2,LM0; hence thesis ; end; EX: ex x,y be object st x in Tokyoite & y in Tokyoite & x <> y & Numberofhair.x = Numberofhair.y proof assume A1: not ( ex x,y be object st x in Tokyoite & y in Tokyoite & x <> y & Numberofhair.x = Numberofhair.y ) ; then A2: for x,y be object st x in Tokyoite & y in Tokyoite & x <> y holds Numberofhair.x <> Numberofhair.y ; A3: dom Numberofhair = Tokyoite by FUNCT_2:def 1; then for x,y be object st x in dom Numberofhair & y in dom Numberofhair & Numberofhair.x = Numberofhair.y holds x = y by A2; then Numberofhair is one-to-one by FUNCT_1:def 4; then card (dom Numberofhair) = card (rng Numberofhair) by CARD_1:70; then card (Tokyoite) = card (rng (Numberofhair)) by A3; hence contradiction by LM4; end; end;
帰納的推論
以下の帰納的推論の誤りの原因と帰納的推論限界はどこにあるのか述べること.
ある過多な飲酒僻がある人が,友人に自分が泥酔する原因はアルコールにあるのではないと主張し,自身の体を使って以下の実験を行った.
- 日曜 ウイスキーの水割りを多量に飲用した
- 月曜 ウォッカのオンザロックを多量に飲用した
- 火曜 ブランデーと炭酸水を多量に飲用した
- 水曜 ワインとミネラルウォーターを多量に飲用した
- 木曜 ジントニックと炭酸水を多量に飲用した
- 金曜 テキーラとミネラルウォーターを多量に飲用した
- 土曜 リキュールとミネラルウォーターを多量に飲用した
当然ながら,この人物は1週間毎日泥酔した.そして,日曜日に自分が飲用したものには全て水が共通している. よって自分が泥酔する原因は水であると主張した.