John, using your new axiom framework, with help from Freek, I ported my Tarski geometry Mizar code up to Gupta's theorem (code below). Thanks again for your three frameworks! Can you look at the top of my code? I could not see how to define the predicates ORDERED a,b,c,d x on_line a,b I failed to modify your `parse_as_infix' defs of === and cong.
BTW Bob Bruner, the mathematician at Wayne State in my subject homotopy theory who got me interested in HOL Light, wrote this: I just read Harrison's talk and the first thing that comes to my mind is to get started on algebraic topology in a systematic way. Let's forget proving just the big theorems and build the whole thing. I wish I had the time to say I am volunteering, but I cannot right now. Of course, I guess you first need to build some algebra and either topological spaces or simplicial sets. -- Best, Bill (* ================================================================= *) (* HOL Light Tarski geometry axiomatic proofs up to Gupta's theorem. *) (* ================================================================= *) (* Proof assistants like HOL Light can be used to help teach rigorous axiomatic geometry in high school using Hilbert's axioms, and introduce students to the world of formal proofs, which should become a hot area in debugging computer software. This is a port, mostly due to John Harrison, of Mizar code, which was heavily influenced by Julien Narboux's Coq pseudo-code http://dpt-info.u-strasbg.fr/~narboux/tarski.html and Wojciech A. Trybulec's incsp_1.miz in the MML library on axioms of incidence geometry. We partially prove the theorem of the 1983 book Metamathematische Methoden in der Geometrie by Schwabhäuser, Szmielew, and Tarski, that Tarski's (extremely weak!) plane geometry axioms imply Hilbert's axioms. We get about as far as Narboux, with Gupta's amazing proof which implies Hilbert's axiom I1 that two points determine a line. Thanks to Mizar folks who wrote an influential language I was able to learn, Freek Wiedijk, who wrote the miz3 port of Mizar to HOL Light, and especially John Harrison, who came up with the entire framework of porting my axiomatic proofs to HOL Light. *) new_type("point",0);; parse_as_infix("===",(12, "right"));; parse_as_infix("cong",(12, "right"));; (* parse_as_infix("is_ordered",(12, "right"));; *) new_constant("===",`:point#point->point#point->bool`);; new_constant("Between",`:point#point#point->bool`);; let cong_DEF = new_definition `a,b,c cong x,y,z <=> a,b === x,y /\ a,c === x,z /\ b,c === y,z`;; let is_ordered_DEF = new_definition `is_ordered (a,b,c,d) <=> Between (a,b,c) /\ Between (a,b,d) /\ Between (a,c,d) /\ Between (b,c,d)`;; (* I want to define is_ordered as a postfix operator, but didn't know how to do it. So I tried to make it a prefix, but this didn't work: parse_as_prefix("ORDERED");; let ORDERED_DEF = new_definition `ORDERED a,b,c,d <=> Between (a,b,c) /\ Between (a,b,d) /\ Between (a,c,d) /\ Between (b,c,d)`;; *) let Collinear_DEF = new_definition `Collinear(a,b,c) <=> Between (a,b,c) \/ Between (b,c,a) \/ Between (c,a,b)`;; (* this doesn't work either parse_as_infix("on_line",(12, "right"));; let Line_DEF = new_definition `x on_line a,b <=> ~(a = b) ∧ (Between (a,b,x) \/ Between (b,x,a) \/ Between (x,a,b))`;; *) (* ------------------------------------------------------------------------- *) (* The axioms. *) (* ------------------------------------------------------------------------- *) let A1 = new_axiom `!a b. a,b === b,a`;; let A2 = new_axiom `!a b p q r s. a,b === p,q /\ a,b === r,s ==> p,q === r,s`;; let A3 = new_axiom `!a b c. a,b === c,c ==> a = b`;; let A4 = new_axiom `!a q b c. ?x. Between(q,a,x) /\ a,x === b,c`;; let A5 = new_axiom `!a b c x a' b' c' x'. ~(a = b) /\ a,b,c cong a',b',c' /\ Between(a,b,x) /\ Between(a',b',x') /\ b,x === b',x' ==> c,x === c',x'`;; let A6 = new_axiom `!a b. Between(a,b,a) ==> a = b`;; let A7 = new_axiom `!a b p q z. Between(a,p,z) /\ Between(b,q,z) ==> ?x. Between(p,x,b) /\ Between(q,x,a)`;; (* A4 is the Segment Construction axiom, A5 is the SAS axiom and A7 is the Inner Pasch axiom. There are 4 more axioms we're not using yet: there exist 3 non-collinear points; 3 points equidistant from 2 distinct points are collinear; Euclid's parallel postulate; a first order version of Hilbert's Dedekind Cuts axiom. We shall say we apply SAS to a+cbx and a'+c'b'x'. Normally one applies SAS by showing cb = c'b' bx = b'x' (which we assume) and angle cbx cong angle c'b'x'. One might prove the angle congruence by showing that the triangles abc /\ a'b'c' were congruent by SSS (which we also assume) and then apply the theorem that complements of congruent angles are congruent. Hence Tarski's axiom. *) (* ------------------------------------------------------------------------- *) (* Now Mizarlight versions of the actual proofs. *) (* ------------------------------------------------------------------------- *) #load "unix.cma";; loadt "miz3/miz3.ml";; horizon := 0;; let EquivReflexive = thm `; !a b. a,b === a,b proof let a b be point; b,a === a,b by A1; qed by -, A2`;; let EquivSymmetric = thm `; !a b c d. a,b === c,d ==> c,d === a,b proof let a b c d be point; assume a,b === c,d [1]; a,b === a,b by EquivReflexive; qed by -, 1, A2`;; let EquivTransitive = thm `; !a b p q r s . a,b === p,q /\ p,q === r,s ==> a,b === r,s proof let a b p q r s be point; assume a,b === p,q [H1]; assume p,q === r,s [H2]; p,q === a,b by H1, EquivSymmetric; qed by -, H2, A2`;; let Baaa_THM = thm `; !a b. Between (a,a,a) /\ a,a === b,b proof let a b be point; consider x such that Between (a,a,x) /\ a,x === b,b [X1] by A4; a = x by -, A3; qed by -, X1`;; let Bqaa_THM = thm `; !a q. Between(q,a,a) proof let a q be point; consider x such that Between(q,a,x) /\ a,x === a,a [X1] by A4; a = x by -, A3; qed by -, X1`;; let C1_THM = thm `; let a b x y be point; assume ~(a = b) [H1]; assume Between (a,b,x) [H2]; assume Between (a,b,y) [H3]; assume b,x === b,y [H4]; thus y = x proof a,b === a,b /\ a,y === a,y /\ b,y === b,y by EquivReflexive; a,b,y cong a,b,y by -, cong_DEF; y,x === y,y by -, H1, H2, H3, H4, A5; qed by -, A3`;; let Bsymmetry_THM = thm `; let a p z be point; thus Between (a,p,z) ==> Between (z,p,a) proof assume Between (a,p,z) [H1]; Between (p,z,z) by Bqaa_THM; consider x such that Between (p,x,p) /\ Between (z,x,a) [X1] by -, H1, A7; x = p by -, A6; qed by -, X1`;; let Baaq_THM = thm `; let a q be point; thus Between (a,a,q) proof Between (q,a,a) by Bqaa_THM; qed by -, Bsymmetry_THM`;; let BEquality_THM = thm `; let a b c be point; thus Between (a,b,c) /\ Between (b,a,c) ==> a = b proof assume Between (a,b,c) [H1]; assume Between (b,a,c); ? x . Between (b,x,b) /\ Between (a,x,a) by -, H1, A7; consider x such that Between (b,x,b) /\ Between (a,x,a) [X1] by -; b = x by X1, A6; Between (a,b,a) by -, X1; qed by -, A6`;; let B124and234then123_THM = thm `; let a b c d be point; assume Between (a,b,d) [H1]; assume Between (b,c,d) [H2]; thus Between (a,b,c) proof ? x . Between (b,x,b) /\ Between (c,x,a) by H1, H2, A7; consider x such that Between (b,x,b) /\ Between (c,x,a) [X1] by -; b = x by X1, A6; Between (c,b,a) by -, X1; qed by -, Bsymmetry_THM`;; let BTransitivity_THM = thm `; let a b c d be point; assume ~(b = c) [H1]; assume Between (a,b,c) [H2]; assume Between (b,c,d) [H3]; thus Between (a,c,d) proof consider x such that Between (a,c,x) /\ c,x === c,d [X1] by A4; Between (x,c,a) [X2] by -, Bsymmetry_THM; Between (c,b,a) by H2, Bsymmetry_THM; Between (x,c,b) by -, X2, B124and234then123_THM; Between (b,c,x) by -, Bsymmetry_THM; x = d by -, H1, H3, X1, C1_THM; qed by -, X1`;; let BTransitivityOrdered_THM = thm `; let a b c d be point; assume ~(b = c) [H1]; assume Between (a,b,c) [H2]; assume Between (b,c,d) [H3]; thus is_ordered (a,b,c,d) proof Between (a,c,d) [X1] by H1, H2, H3, BTransitivity_THM; Between (d,c,b) [X2] by H3, Bsymmetry_THM; Between (c,b,a) by -, H2, Bsymmetry_THM; Between (d,b,a) by -, H1, X2, BTransitivity_THM; Between (a,b,d) by -, Bsymmetry_THM; qed by H2, -, X1, H3, is_ordered_DEF`;; (* let BTransitivityOrdered_THM = thm `; ! a b c d . ~(b = c) /\ Between (a,b,c) /\ Between (b,c,d) ==> ORDERED a,b,c,d proof let a b c d be point; assume ~(b = c) [H1]; assume Between (a,b,c) [H2]; assume Between (b,c,d) [H3]; Between (a,c,d) [X1] by H1, H2, H3, BTransitivity_THM; Between (d,c,b) [X2] by H3, Bsymmetry_THM; Between (c,b,a) by -, H2, Bsymmetry_THM; Between (d,b,a) by -, H1, X2, BTransitivity_THM; Between (a,b,d) by -, Bsymmetry_THM; thus ORDERED a,b,c,d by H2, -, X1, H3, ORDERED_DEF; end`;; *) let B124and234Ordered_THM = thm `; let a b c d be point; assume Between (a,b,d) [H1]; assume Between (b,c,d) [H2]; thus is_ordered (a,b,c,d) proof cases; suppose b = c [P1]; Between (a,b,c) [P2] by -, Bqaa_THM; Between (a,c,d) by P1, H1; qed by P2, H1, -, H2, is_ordered_DEF; suppose ~(b = c) [Q1]; Between (a,b,c) by H1, H2, B124and234then123_THM; qed by -, Q1, H2, BTransitivityOrdered_THM; end`;; let SegmentAddition_THM = thm `; let a b c a' b' c' be point; assume Between (a,b,c) [H1]; assume Between (a',b',c') [H2]; assume a,b === a',b' [H3]; assume b,c === b',c' [H4]; thus a,c === a',c' proof cases; suppose a = b [Y1]; a,a === a',b' by H3, Y1; a',b' === a,a by -, EquivSymmetric; a' = b' by -, A3; qed by -, H4, Y1; suppose ~(a = b) [Z1]; b,a === a,b by A1; b,a === a',b' [Z2] by -, H3, EquivTransitive; a',b' === b',a' by A1; b,a === b',a' [Z3] by -, Z2, EquivTransitive; a,a === a',a' by Baaa_THM; a,b,a cong a',b',a' by -, H3, Z3, cong_DEF; qed by -, Z1, H1, H2, H4, A5; end`;; let CongruenceDoubleSymmetry_THM = thm `; let a b c d be point; assume a,b === c,d [H1]; thus b,a === d,c proof b,a === a,b /\ c,d === d,c [X1] by H1, A1; a,b === d,c by H1, X1, EquivTransitive; qed by -, X1, EquivTransitive`;; let C1prime_THM = thm `; let a b x y be point; assume ~(a = b) [H1]; assume Between (a,b,x) [H2]; assume Between (a,b,y) [H3]; assume a,x === a,y [H4]; thus x = y proof consider m such that Between (b,a,m) /\ a,m === a,b [X1] by A4; Between (m,a,b) [X2] by X1, Bsymmetry_THM; ~(m = a) [X3] by X1, EquivSymmetric, A3, H1; is_ordered (m,a,b,x) by H1, X2, H2, BTransitivityOrdered_THM; Between (m,a,x) [X4] by -, is_ordered_DEF; is_ordered (m,a,b,y) by H1, X2, H3, BTransitivityOrdered_THM; Between (m,a,y) by -, is_ordered_DEF; qed by -, X3, X4, H4, C1_THM`;; let SegmentSubtraction_THM = thm `; let a b c a' b' c' be point; assume Between (a,b,c) [H1]; assume Between (a',b',c') [H2]; assume a,b === a',b' [H3]; assume a,c === a',c' [H4]; thus b,c === b',c' proof cases; suppose a = b [Y1]; a,a === a',b' by -, H3; a',b' === a,a by -, EquivSymmetric; a' = b' by -, A3; qed by -, H4, Y1; suppose ~(a = b) [Z1]; consider x such that Between (a,b,x) /\ b,x === b',c' [Z2] by A4; a,x === a',c' [Z3] by Z2, H2, H3, SegmentAddition_THM; a',c' === a,c by H4, EquivSymmetric; a,x === a,c by -, Z3, EquivTransitive; x = c by -, Z1, Z2, H1, C1prime_THM; qed by -, Z2; end`;; let EasyAngleTransport_THM = thm `; let a O b be point; assume ~(O = a) [H1]; thus ? x y . Between (b,O,x) /\ Between (a,O,y) /\ x,y,O cong a,b,O proof consider x such that Between (b,O,x) /\ O,x === O,a [X2] by A4; x,O === a,O [X3] by -, CongruenceDoubleSymmetry_THM; a,O === x,O [X4] by -, EquivSymmetric; a,x === x,a by A1; a,O,x cong x,O,a [X5] by X4, -, X2, cong_DEF; consider y such that Between (a,O,y) /\ O,y === O,b [X6] by A4; Between (x,O,b) by X2 ,Bsymmetry_THM; x,y === a,b [X7] by H1, X5, X6, -, A5; y,O === b,O by X6, CongruenceDoubleSymmetry_THM; x,y,O cong a,b,O by X7, X3, -, cong_DEF; qed by X2, X6, -`;; let B123and134Ordered_THM = thm `; let a b c d be point; assume Between (a,b,c) [H1]; assume Between (a,c,d) [H2]; thus is_ordered (a,b,c,d) proof Between (d,c,a) /\ Between (c,b,a) by H2, H1, Bsymmetry_THM; is_ordered (d,c,b,a) by -, B124and234Ordered_THM; Between (d,b,a) /\ Between (d,c,b) by -, is_ordered_DEF; Between (a,b,d) /\ Between (b,c,d) by -, Bsymmetry_THM; thus is_ordered (a,b,c,d) by -, H1, H2, is_ordered_DEF; end`;; let BextendToLine_THM = thm `; let a b c d be point; assume ~(a = b) [H1]; assume Between (a,b,c) [H2]; assume Between (a,b,d) [H3]; thus ? x . is_ordered (a,b,c,x) /\ is_ordered (a,b,d,x) proof consider u such that Between (a,c,u) /\ c,u === b,d [X1] by A4; is_ordered (a,b,c,u) [X2] by H2, X1, B123and134Ordered_THM; Between (b,c,u) by X2, is_ordered_DEF; Between (u,c,b) [X3] by -, Bsymmetry_THM; u,c === c,u by A1; u,c === b,d [X4] by -, X1, EquivTransitive; Between (a,b,u) [X5] by X2, is_ordered_DEF; consider x such that Between (a,d,x) /\ d,x === b,c [Y1] by A4; is_ordered (a,b,d,x) [Y2] by H3, Y1, B123and134Ordered_THM; Between (b,d,x) [Y3] by -, is_ordered_DEF; b,c === d,x [Y4] by Y1, EquivSymmetric; c,b === b,c by A1; c,b === d,x [Y5] by -, Y4, EquivTransitive; Between (a,b,x) [Y6] by Y2, is_ordered_DEF; u,b === b,x [X6] by X3, Y3, X4, Y5, SegmentAddition_THM; b,u === u,b by A1; b,u === b,x by -, X6, EquivTransitive; u = x by -, H1, X5, Y6, C1_THM; qed by -, X2, Y2`;; let GuptaEasy_THM = thm `; let a b c d be point; assume ~(a = b) [H1]; assume Between (a,b,c) [H2]; assume Between (a,b,d) [H3]; assume ~(b = c) [H4]; assume ~(b = d) [H5]; thus ~ Between (c,b,d) proof ~ Between (c,b,d) \/ Between (c,b,d) [1]; cases by 1; suppose ~ Between (c,b,d); qed by -; suppose Between (c,b,d) [H6]; ? x . is_ordered (a,b,c,x) /\ is_ordered (a,b,d,x) by H1, H2, H3, BextendToLine_THM; consider x such that is_ordered (a,b,c,x) /\ is_ordered (a,b,d,x) [X1] by -; Between (b,d,x) by X1, is_ordered_DEF; is_ordered (c,b,d,x) by -, H5, H6, BTransitivityOrdered_THM; Between (b,c,x) /\ Between (c,b,x) by -, X1, is_ordered_DEF; b = c [X2] by -, BEquality_THM; F by -, H4, X2; qed by -; end`;; (* The next result is like SAS: there are 5 pairs of segments 4 equivalent. We say we apply Inner5Segments to abc-x and a'b'c'-x' *) let Inner5Segments_THM = thm `; let a b c x a' b' c' x' be point; assume a,b,c cong a',b',c' [H1]; assume Between (a,x,c) [H2]; assume Between (a',x',c') [H3]; assume c,x === c',x' [H4]; thus b,x === b',x' proof a,b === a',b' /\ a,c === a',c' /\ b,c === b',c' [X1] by H1, cong_DEF; cases; suppose x = c [Case1]; c',x' === c,c by H4, Case1, EquivSymmetric; x' = c' by -, A3; qed by -, Case1, X1; suppose ~(x = c) [Case2]; ~(a = c) [X2] by H2, A6, Case2; consider y such that Between (a,c,y) /\ c,y === a,c [X3] by A4; consider y' such that Between (a',c',y') /\ c',y' === a,c [X4] by A4; a,c === c',y' by X4, EquivSymmetric; c,y === c',y' [X5] by -, X3, EquivTransitive; c,b === c',b' [X6] by X1, CongruenceDoubleSymmetry_THM; a,c,b cong a',c',b' by cong_DEF, X1, X6; b,y === b',y' [X7] by -, X2, X3, X4, X5, A5; ~(y = c) [X8] by X3, EquivSymmetric, A3, X2; Between (y,c,a) /\ Between (c,x,a) by X3, H2, Bsymmetry_THM; Between (y,c,x) [X9] by -, B124and234then123_THM; Between (y',c',a') /\ Between (c',x',a') by -, X4, H3, Bsymmetry_THM; Between (y',c',x') [X10] by -, B124and234then123_THM; y,c === y',c' /\ y,b === y',b' by X5, X7, CongruenceDoubleSymmetry_THM; y,c,b cong y',c',b' by -, cong_DEF, X6; qed by -, X8, X9, X10, H4, A5; end`;; let RhombusDiagBisect_THM = thm `; let b c d c' d' be point; assume Between (b,c,d') [H1]; assume Between (b,d,c') [H2]; assume c,d' === c,d [H3]; assume d,c' === c,d [H4]; assume d',c' === c,d [H5]; thus ? e . Between (c,e,c') /\ Between (d,e,d') /\ c,e === c',e /\ d,e === d',e proof Between (d',c,b) /\ Between (c',d,b) [X1] by H1, H2, Bsymmetry_THM; consider e such that Between (c,e,c') /\ Between (d,e,d') [X2] by X1, A7; c,d === c,d' [X3] by H3, EquivSymmetric; c,c' === c,c' [X4] by EquivReflexive; c,d === d',c' by H5, EquivSymmetric; d,c' === d',c' by -, H4, EquivTransitive; c,d,c' cong c,d',c' by -, X3, X4, cong_DEF; d,e === d',e [X5] by -, X2, EquivReflexive, Inner5Segments_THM; d,c === c,d [X6] by A1; c,d === d,c' by H4, EquivSymmetric; d,c === d,c' [X7] by -, X6, EquivTransitive; d,d' === d,d' [X8] by EquivReflexive; c,d === d',c' [X9] by H5, EquivSymmetric; d',c' === c',d' by A1; c,d === c',d' by -, X9, EquivTransitive; c,d' === c',d' [X10] by -, H3, EquivTransitive; d,d' === d,d' by EquivReflexive; d,c,d' cong d,c',d' by -, X7, X8, X10, cong_DEF; c,e === c',e by -, X2, EquivReflexive, Inner5Segments_THM; qed by -, X2, X5`;; (* In proof below, Apply SAS to p+crq /\ d'+ced *) let FlatNormal_THM = thm `; let a b c d d' e be point; assume Between (b,c,d') [H1]; assume Between (d,e,d') [H2]; assume c,d' === c,d [H3]; assume d,e === d',e [H4]; assume ~(c = d) [H5]; assume ~(e = d) [H6]; thus ? p r q . Between (p,r,q) /\ Between (r,c,d') /\ Between (e,c,p) /\ r,c,p cong r,c,q /\ r,c === e,c /\ p,r === d,e proof ~(c = d') by H5, H3, EquivSymmetric, A3; consider p r such that Between (e,c,p) /\ Between (d',c,r) /\ p,r,c cong d',e,c [X1] by -, EasyAngleTransport_THM; p,r === d',e /\ p,c === d',c /\ r,c === e,c [X2] by -, X1, cong_DEF; d',e === d,e by H4, EquivSymmetric; p,r === d,e [X3] by -, X2, EquivTransitive; ~(p = r) [X4] by -, EquivSymmetric, H6, A3; consider q such that Between (p,r,q) /\ r,q === e,d [X5] by A4; Between (d',e,d) [X6] by H2, Bsymmetry_THM; c,p === c,d' by -, X2, CongruenceDoubleSymmetry_THM; c,p === c,d [X7] by -, H3, EquivTransitive; c,q=== c,d by X4, X1, X5, X6, A5; c,d=== c,q by -, EquivSymmetric; c,p=== c,q [X8] by -, X7, EquivTransitive; r,c=== r,c [X9] by EquivReflexive; r,p=== e,d [X10] by X3, CongruenceDoubleSymmetry_THM; e,d=== r,q by X5, EquivSymmetric; r,p=== r,q by -, X10, EquivTransitive; r,c,p cong r,c,q [X11] by -, X9, X8, cong_DEF; Between (r,c,d') [X12] by X1, Bsymmetry_THM; qed by -, X5, X11, X12, X2, X1, X3`;; (* a /\ b are equidistant from p /\ q. Apply SAS to a+pbc /\ a+qbc. *) let EqDist2PointsBetween_THM = thm `; let a b c p q be point; assume ~(a = b) [H1]; assume Between (a,b,c) [H2]; assume a,p === a,q /\ b,p === b,q [H3]; thus c,p === c,q proof a,b === a,b /\ b,c === b,c [X1] by EquivReflexive; a,b,p cong a,b,q by -, H3, cong_DEF; p,c === q,c by H1, -, H2, X1, A5; qed by -, CongruenceDoubleSymmetry_THM`;; (* a and c are equidistant from p and q. Apply Inner5Segments to apb-x /\ aqb-x. *) let EqDist2PointsInnerBetween_THM = thm `; let a x c p q be point; assume Between (a,x,c) [H1]; assume a,p === a,q /\ c,p === c,q [H2]; thus x,p === x,q proof a,c === a,c /\ c,x === c,x [X1] by EquivReflexive; p,c === q,c by H2, CongruenceDoubleSymmetry_THM; a,p,c cong a,q,c by -, H2, X1, cong_DEF; p,x === q,x by -, H1, X1, Inner5Segments_THM; thus x,p === x,q by -, CongruenceDoubleSymmetry_THM; end`;; (* In the proof below (after X8), we prove a stronger result than BextendToLine_THM with much the same proof. We find u /\ b' with essentially a,b,c,d',u and a b,d,c',b' ordered 5-tuples with d'u === db /\ cb' === bc. *) (* Show (after Y13) c'd' === cd by applying SAS to b+c'cd /\ b'+cc'd. *) (* (before Z4) c,d',c',d is a ``flat'' rhombus. The diagonals bisect each other: *) (* (after W2) r and c are equidistant from p and q, r <> c, Between r,c,d', thus also d' *) (* (after W3) c and d' are equidistant from p and q, c <> d', Between c,d',b', thus also b'. *) (* (after W4) d' and c are equidistant from p and q, d' <> c, Between d',c,b, thus also b. *) (* (after W5) b and b' are equidistant from p and q, Between b,c',b, thus also c'. *) (* (after W7) c' and c are equidistant from p and q, c' <> c, Between c',c,p, thus also p. *) (* (after p,p === p,q) Now we deduce a contradiction from p = q. *) let Gupta_THM = thm `; let a b c d be point; assume ~(a = b) [H1]; assume Between (a,b,c) [H2]; assume Between (a,b,d) [H3]; thus Between (b,d,c) \/ Between (b,c,d) proof (b = c \/ b = d \/ c = d) \/ (~(b = c) /\ ~(b = d) /\ ~(c = d)) [1]; cases by 1; suppose b = c \/ b = d \/ c = d; Between (b,d,c) \/ Between (b,c,d) by -, Baaq_THM, Bqaa_THM; qed by -; suppose ~(b = c) /\ ~(b = d) /\ ~(c = d) [H4]; assume ~ (Between (b,d,c)) [H5]; consider d' such that Between (a,c,d') /\ c,d' === c,d [X1] by A4; consider c' such that Between (a,d,c') /\ d,c' === c,d [X2] by A4; is_ordered (a,b,c,d') by H2, X1, B123and134Ordered_THM; Between (a,b,d') /\ Between (b,c,d') [X3] by -, is_ordered_DEF; is_ordered (a,b,d,c') by H3, X2, B123and134Ordered_THM; Between (a,b,c') /\ Between (b,d,c') [X4] by -, is_ordered_DEF; ~(c = d') [X5] by X1, H4, A3, EquivSymmetric; ~(d = c') [X6] by X2, H4, A3, EquivSymmetric; ~(b = d') [X7] by X3, H4, A6; ~(b = c') [X8] by X4, H4, A6; consider u such that Between (c,d',u) /\ d',u === b,d [Y1] by A4; is_ordered (b,c,d',u) by X5, X3, Y1, BTransitivityOrdered_THM; Between (b,c,u) /\ Between (b,d',u) [Y2] by -, is_ordered_DEF; consider b' such that Between (d,c',b') /\ c',b' === b,c [Y3] by A4; is_ordered (b,d,c',b') by X6, X4, Y3, BTransitivityOrdered_THM; Between (b,d,b') /\ Between (b,c',b') [Y4] by -, is_ordered_DEF; Between (c',d,b) [Y5] by X4, Bsymmetry_THM; d,c' === c',d /\ b,d === d,b [Y6] by A1; c,d === d,c' by X2, EquivSymmetric; c,d' === d,c' by -, X1, EquivTransitive; c,d' === c',d [Y7] by -, Y6, EquivTransitive; d',u === d,b by Y1, Y6, EquivTransitive; c,u === c',b [Y8] by -, Y1, Y5, Y7, SegmentAddition_THM; c',b' === b',c' /\ b',b === b,b' [Y9] by A1; b,c === c',b' by Y3, EquivSymmetric; b,c === b',c' [Y10] by -, Y9, EquivTransitive; Between (b',c',b) by Y4, Bsymmetry_THM; b,u === b',b by -, Y2, Y10, Y8, SegmentAddition_THM; b,u === b,b' [Y11] by -, Y9, EquivTransitive; is_ordered (a,b,d',u) [Y12] by X7, X3, Y2, BTransitivityOrdered_THM; is_ordered (a,b,c',b') by X8, X4, Y4, BTransitivityOrdered_THM; Between (a,b,u) /\ Between (a,b,b') by -, Y12, is_ordered_DEF; u = b' [Y13] by -, H1, Y11, C1_THM; c',b === c,b' by Y13, Y8, EquivSymmetric; b,c' === b',c [Z1] by -, CongruenceDoubleSymmetry_THM; c,c' === c',c by A1; b,c,c' cong b',c',c [Z2] by -, Y10, Z1, cong_DEF; Between (b',c',d) by Y3, Bsymmetry_THM; c',d' === c,d [Z3] by -, H4, Z2, X3, Y7, A5; d',c' === c',d' by A1; d',c' === c,d by -, Z3, EquivTransitive; consider e such that Between (c,e,c') /\ Between (d,e,d') /\ c,e === c',e /\ d,e === d',e [Z4] by -, X3, X4, X1, X2, RhombusDiagBisect_THM; ~(e = c) [U1] proof cases; suppose ~(e = c); qed by -; suppose e = c [U2]; c' = e by U2, Z4, EquivSymmetric, A3; c' = c by -, U2; Between (b,d,c) [U3] by -, X4; F by -, U3, H5; qed by -; end; e = d [V1] proof cases; suppose e = d; qed by -; suppose ~(e = d) [V2]; consider p r q such that Between (p,r,q) /\ Between (r,c,d') /\ Between (e,c,p) /\ r,c,p cong r,c,q /\ r,c === e,c /\ p,r === d,e [W1] by X3, Z4, X1, H4, V2, FlatNormal_THM; r,p === r,q /\ c,p === c,q [W2] by W1, cong_DEF; ~(c = r) by W1, U1, EquivSymmetric, A3; d',p === d',q [W3] by -, W1, W2, EqDist2PointsBetween_THM; Between (c,d',b') by Y1, Y13; b',p === b',q [W4] by -, X5, W2, W3, EqDist2PointsBetween_THM; Between (d',c,b) by X3, Bsymmetry_THM; b,p === b,q [W5] by -, X5, W3, W2, EqDist2PointsBetween_THM; c',p === c',q [W7]by Y4, W4, W5, EqDist2PointsInnerBetween_THM; Between (c',e,c) by Z4, Bsymmetry_THM; is_ordered (c',e,c,p) by -, U1, W1, BTransitivityOrdered_THM; Between (c',c,p) [W8] by -, is_ordered_DEF; ~(c' = c) by Z4, U1, A6; p,p === p,q by -, W8, W7, W2, EqDist2PointsBetween_THM; q = p by -, EquivSymmetric, A3; p = r by -, W1, A6; e = d [W9] by -, W1, EquivSymmetric, A3; F by -, W9, V2; qed by -; end; d' = e by V1, Z4, EquivSymmetric, A3; d' = d by -, V1; Between (b,c,d) by -, X3; qed by -; end`;; ------------------------------------------------------------------------------ Live Security Virtual Conference Exclusive live event will cover all the ways today's security and threat landscape has changed and how IT managers can respond. Discussions will include endpoint security, mobile security and the latest in malware threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/ _______________________________________________ hol-info mailing list hol-info@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/hol-info
