John, I had a blast continuing your port of my Tarksi geometry code,
included below. I'm about 1/3 done (below), and I have a serious
problem, which I think underscores the lack of Mizar dox.
I can't understand what's wrong with this proof below. This is the
only thm I wrote where the conclusion is there-exists statement, and
I'm guessing that's the problem.
********************************
let EasyAngleTransport_THM = thm `;
! a o b . (~(o = a) ==>
(? x y . (Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o))))
proof
let a o b be point;
assume ~(o = a) [X1];
? x . Between (b,o,x) /\ o,x === o,a by A4;
consider x such that
Between (b,o,x) /\ o,x === o,a [X2] by -;
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;
? y . Between (a,o,y) /\ o,y === o,b by A4;
consider y such that
Between (a,o,y) /\ o,y === o,b [X6] by -;
Between (x,o,b) by X2 ,Bsymmetry_THM;
x,y === a,b [X7] by X1, X5, X6, -, A5;
y,o === b,o by X6, CongruenceDoubleSymmetry_THM;
x,y,o cong a,b,o by X7, X3, -, cong_DEF;
thus Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o) by X2, X6, -;
end`;;
The to-me incomprehensible Mizar_error:
(`;
! a o b . (~(o = a) ==>
(? x y . (Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o))))
:: #8
:: 8: syntax or type error hol
proof
let a o b be point;
assume ~(o = a) [X1];
:: #8
? x . Between (b,o,x) /\ o,x === o,a by A4;
:: #8
consider x such that
Between (b,o,x) /\ o,x === o,a [X2] by -;
:: #8
x,o === a,o [X3] by -, CongruenceDoubleSymmetry_THM;
:: #8
a,o === x,o [X4] by -, EquivSymmetric;
:: #8
a,x === x,a by A1;
:: #7
:: 7: unbound free variables hol
a,o,x cong x,o,a [X5] by X4, -, X2, cong_DEF;
:: #8
? y . Between (a,o,y) /\ o,y === o,b by A4;
:: #8
consider y such that
Between (a,o,y) /\ o,y === o,b [X6] by -;
:: #8
Between (x,o,b) by X2 ,Bsymmetry_THM;
:: #8
x,y === a,b [X7] by X1, X5, X6, -, A5;
:: #7
y,o === b,o by X6, CongruenceDoubleSymmetry_THM;
:: #8
x,y,o cong a,b,o by X7, X3, -, cong_DEF;
:: #8
thus Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o) by X2, X6,
-;
::
#8
end
;`,
(15, 0, 0)).
**********************************************************
--
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)`;;
*)
(* ------------------------------------------------------------------------- *)
(* 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 : point. 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 [1];
assume p,q === r,s [2];
p,q === a,b by 1, EquivSymmetric;
qed by -, 2, A2`;;
let Baaa_THM = thm `;
!a b. Between (a,a,a) /\ a,a === b,b
proof
let a b be point;
?x. Between (a,a,x) /\ a,x === b,b by A4;
consider x such that Between (a,a,x) /\ a,x === b,b [1] by - ;
a = x by 1, A3;
qed by -, 1`;;
let Bqaa_THM = thm `;
!a q. Between(q,a,a)
proof
let a q be point;
? x : point . Between(q,a,x) /\ a,x === a,a by A4;
consider x such that Between(q,a,x) /\ a,x === a,a [1] by -;
a = x by 1, A3;
qed by -, 1`;;
let C1_THM = thm `;
!a b x y . ~(a = b) /\ Between (a,b,x) /\ Between (a,b,y) /\ b,x === b,y
==> y = x
proof
let a b x y be point;
assume ~(a = b) [1];
assume Between (a,b,x) [2];
assume Between (a,b,y) [3];
assume b,x === b,y [4];
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 -, 1, 2, 3, 4, A5;
qed by -, A3`;;
let Bsymmetry_THM = thm `;
! a p z .
Between (a,p,z) ==> Between (z,p,a)
proof
let a p z be point;
assume Between (a,p,z) [H1];
Between (p,z,z) by Bqaa_THM;
?x . Between (p,x,p) /\ Between (z,x,a) by -, H1, A7;
consider x such that
Between (p,x,p) /\ Between (z,x,a) [X1] by -;
x = p by -, A6;
qed by -, X1`;;
let Baaq_THM = thm `;
! a q . Between (a,a,q)
proof
let a q be point;
Between (q,a,a) by Bqaa_THM;
qed by -, Bsymmetry_THM`;;
let BEquality_THM = thm `;
! a b c . Between (a,b,c) /\ Between (b,a,c) ==> a = b
proof
let a b c be point;
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 `;
! a b c d .
Between (a,b,d) /\ Between (b,c,d) ==> Between (a,b,c)
proof
let a b c d be point;
assume Between (a,b,d) [H1];
assume Between (b,c,d);
? x . Between (b,x,b) /\ Between (c,x,a) by -, H1, 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 `;
! a b c d .
~(b = c) /\ Between (a,b,c) /\ Between (b,c,d) ==> Between (a,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];
? x . Between (a,c,x) /\ c,x === c,d by A4;
consider x such that
Between (a,c,x) /\ c,x === c,d [X1] by -;
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 `;
! a b c d .
~(b = c) /\ Between (a,b,c) /\ Between (b,c,d) ==> is_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;
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 `;
! a b c d .
Between (a,b,d) /\ Between (b,c,d) ==> is_ordered (a,b,c,d)
proof
let a b c d be point;
assume Between (a,b,d) [H1];
assume Between (b,c,d) [H2];
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 `;
! a b c a' b' c' .
Between (a,b,c) /\ Between (a',b',c') /\ a,b === a',b' /\ b,c === b',c'
==> a,c === a',c'
proof
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];
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
? m . Between (b,a,m) /\ a,m === a,b by A4;
consider m such that
Between (b,a,m) /\ a,m === a,b [X1] by -;
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`;;
(* this doesn't work:
let EasyAngleTransport_THM = thm `;
! a o b . (~(o = a) ==>
(? x y . (Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o))))
proof
let a o b be point;
assume ~(o = a) [H1];
? x . Between (b,o,x) /\ o,x === o,a by A4;
consider x such that
Between (b,o,x) /\ o,x === o,a [X2] by -;
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;
? y . Between (a,o,y) /\ o,y === o,b by A4;
consider y such that
Between (a,o,y) /\ o,y === o,b [X6] by -;
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;
thus Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o) by X2, X6, -;
end`;;
The first Mizar_error is:
(? x y . (Between (b,o,x) /\ Between (a,o,y) /\ (x,y,o cong a,b,o))))
:: #8
:: 8: syntax or type error hol
This modification doesn't work either:
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
? x . Between (b,o,x) /\ o,x === o,a by A4;
consider x such that
Between (b,o,x) /\ o,x === o,a [X2] by -;
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;
? y . Between (a,o,y) /\ o,y === o,b by A4;
consider y such that
Between (a,o,y) /\ o,y === o,b [X6] by -;
Between (x,o,b) by X2, Bsymmetry_THM;
x,y === a,b [X7] by H1, X5, X6, -, 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 Between (a,b,c) /\ Between (a,c,d) ==> 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
? u . Between (a,c,u) /\ c,u === b,d by A4;
consider u such that
Between (a,c,u) /\ c,u === b,d [X1] by -;
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;
? x . Between (a,d,x) /\ d,x === b,c by A4;
consider x such that
Between (a,d,x) /\ d,x === b,c [Y1] by -;
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;
thus thesis by -, X2, Y2;
end`;;
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;
? e . Between (c,e,c') /\ Between (d,e,d') by X1, A7;
consider e such that
Between (c,e,c') /\ Between (d,e,d') [X2] by -;
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;
thus thesis by -, X2, X5;
end`;;
(* can't work without EasyAngleTransport
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;
? p r . Between (e,c,p) /\ Between (d',c,r) /\ p,r,c cong d',e,c by
EasyAngleTransport;
consider p r such that
Between (e,c,p) /\ Between (d',c,r) /\ p,r,c cong d',e,c [X1] by -;
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;
? q . Between (p,r,q) /\ r,q === e,d by A4;
consider q such that
Between (p,r,q) /\ r,q === e,d [X5] by -;
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;
:: Apply SAS to p+crq ∧ d'+ced
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;
thus thesis by X5, X11, X12, X2, X1, X3;
end`;;
*)
(* 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`;;
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