Freek, I just caught myself making a big inference leap that I didn't intend.
The line below
F NOTIN l [notFl] by l_line, Distinct, I1, Collinear_DEF, Fexists,
NOTIN;
is a bug on my part. I meant to write
F NOTIN l [notFl] by l_line, Cl, Collinear_DEF, -, NOTIN;
Miz3 justified my first buggy line with a 547,963 MESON solved at number, and
it took me a while to see how MESON did it. Here's the miz3 proof of an easy
result, Euclid's Proposition I.11: you can push a perpendicular off a line:
let EuclidProp11_THM = thm `;
let A B be point;
assume ~(A = B) [notAB];
thus ? F. Right (angle A B F)
proof
consider l such that
Line l /\ A IN l /\ B IN l [l_line] by notAB, I1;
consider C such that
B IN open (A,C) /\ seg B C === seg B A [ABC] by notAB, SEGMENT,
C1OppositeRay_THM;
C IN l [Cl] by l_line, -, BetweenLinear_THM;
~(A = B) /\ ~(A = C) /\ ~(B = C) /\ Collinear A B C [Distinct] by ABC,
B1';
seg B A === seg B C [BAeqBC] by -, SEGMENT, ABC, C2Symmetric;
consider F such that
~Collinear A F C /\ seg F A === seg F C [Fexists] by Distinct,
IsoscelesExists_THM;
F NOTIN l [notFl] by l_line, Distinct, I1, Collinear_DEF, Fexists,
NOTIN;
~Collinear B F A /\ ~Collinear B F C [BFAncol] by l_line, Cl, Distinct,
I1, Collinear_DEF, -, NOTIN;
~Collinear A B F /\ Angle (angle A B F) [angABF] by -,
CollinearSymmetry_THM, ANGLE;
angle A B F Suppl angle F B C [ABFsuppl] by -, ABC,
SupplementaryAngles_DEF;
~(B = F) by l_line, notFl, NOTIN;
seg B F === seg B F by -, SEGMENT, C2Reflexive;
B,F,A cong B,F,C by BFAncol, -, BAeqBC, Fexists, SSS_THM;
angle A B F === angle F B C by -, TriangleCong_DEF, AngleSymmetry_THM;
qed by angABF, ABFsuppl, -, RightAngle_DEF`;;
All I needed was to say that A, F & C are noncollinear (Fexists), so there
can't be any line containing the three (Collinear_DEF) A & C already belong to
l (l_line, Cl), so if F belongs to l as well, A, F & C would be collinear, and
that's a contradiction. MESON does such proofs by contradiction just fine.
That's the line I should have written:
F NOTIN l [notFl] by l_line, Cl, Collinear_DEF, -, NOTIN;
But I wrote buggily
F NOTIN l [notFl] by l_line, Distinct, I1, Collinear_DEF, Fexists,
NOTIN;
I think I see how miz3 get a proof out of this nonsense I wrote:
A & B are distinct points on l (l_line, Distinct), and so there is only one
line containing A & B (I1). Since
A, B & C are collinear (Distinct), C must belong to this line l as well
(Collinear_DEF), and now we can use the proof I meant above. Maybe this leap
isn't that huge. But I certainly didn't intend it, and it took me a while to
figure out how miz3 could have made sense of it.
--
Best,
Bill
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