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From: Bill Richter <richter@ma...>  20120514 05:05:59

Makarius, here's a miz3 version of an Isabelle proof you asked me to look at, from the Group theory examples in the Isabelle/Isar Reference Manual, p 1819 isabelle.in.tum.de/doc/isarref.pdf which gives a nice proof that left units/identity implies right units. I find the Isabelle proof very hard to read, and I prefer the miz3 proof below. Can you tell me what I'm missing? Can you write this Mizarstyle proof in Isabelle? Maybe it would be nice to be able to mix Isabelle locales and fancy symbols with the miz3 style proofs. (* Paste in these 2 commands: hol_light> ocaml #use "hol.ml";; then paste in the following file*) parse_as_infix("***",(20,"right"));; #load "unix.cma";; loadt "miz3/miz3.ml";; horizon := 0;; let right_inv = thm `; let g be A>bool; let (**) be A>A>A; let i be A>A; let e be A; assume (e IN g) /\ (!x. x IN g ==> i(x) IN g) /\ (!x y. x IN g /\ y IN g ==> x ** y IN g) [1]; assume !x y z. x IN g /\ y IN g /\ z IN g ==> x ** (y ** z) = (x ** y) ** z [Assoc]; assume !x. x IN g ==> e ** x = x [Left_Unit]; assume !x. x IN g ==> i(x) ** x = e [Left_Inv]; thus ! x. x IN g ==> x ** i(x) = e :: x x' = 1(x x') = (1 x) x' = ((x'' x') x) x' = (x'' (x' x)) x' = (x'' 1) x' :: = x'' (1 x') = x'' x' = 1 proof let x be A; assume x IN g [xing]; x ** i(x) = e ** (x ** i(x)) by 1, xing, Left_Unit; x ** i(x) = (e ** x) ** i(x) by 1, xing, , Assoc; x ** i(x) = ((i(i(x)) ** i(x)) ** x) ** i(x) by 1, xing, , Left_Inv; x ** i(x) = (i(i(x)) ** (i(x) ** x)) ** i(x) by 1, xing, , Assoc; x ** i(x) = (i(i(x)) ** e) ** i(x) by 1, xing, , Left_Inv; x ** i(x) = i(i(x)) ** (e ** i(x)) by 1, xing, , Assoc; x ** i(x) = i(i(x)) ** i(x) by 1, xing, , Left_Unit; qed by 1, xing, , Left_Inv`;; (* It works! The output is  !g (**) i e. e IN g /\ (!x. x IN g ==> i x IN g) /\ (!x y. x IN g /\ y IN g ==> x ** y IN g) ==> (!x y z. x IN g /\ y IN g /\ z IN g ==> x ** y ** z = (x ** y) ** z) ==> (!x. x IN g ==> e ** x = x) ==> (!x. x IN g ==> i x ** x = e) ==> (!x. x IN g ==> x ** i x = e) Some discussion: There's no question that the Isabelle proof is easier on the eyes. I checked that I can replace `**' by `*', but you \circ is nicer, and your x^{1} exponent is a lot more readable than i(x). More substantively, it hurts readability to state all the left group theory axioms in the statement of the theorem. But most substantively, the miz3 proof is easier for me to read. A modification of Freek's lagrange1.ml is let group = new_definition `group(g,(**),i,(e:A)) <=> (e IN g) /\ (!x. x IN g ==> i(x) IN g) /\ (!x y. x IN g /\ y IN g ==> x**y IN g) /\ (!x y z. x IN g /\ y IN g /\ z IN g ==> x**(y**z) = (x**y)**z) /\ (!x. x IN g ==> e**x = x) /\ (!x. x IN g ==> x**i(x) = e )`;; That's something like an Isabelle locale, but I don't know how to put labels on the axioms, and Freek did not do so. The clever Isabelle proof uses the facts x'' x' = 1 and x' x = 1 at some points, and uses assoc, left ident & left inverses: x x' = 1(x x') = (1 x) x' = ((x'' x') x) x' = ((x'' (x' x)) x' = ((x'' 1) x' = x'' (1 x') = x'' x' = 1 I can't see how Isabelle justified this proof. I know `...' means the previous right hand side. But Isabelle is doing other things too. Isabelle proved four separate statements S1: x x' = 1(x x') = (1 x) x' S2: (1 x) = (x'' x') x S3: (x'' (x' x)) x' = x'' (1 x') S4: x'' (1 x') = 1 and said finally we're done! Well, why are we done? Isabelle must have combined S3 & S4 to get (x'' (x' x)) x' = 1 and then multiplied S2 on the right by x' to get (1 x) x' = ((x'' x') x) x' combined the last two statements to (1 x) x' = 1 and combined this with S1 to get x x' = 1. I don't see the Isabelle mechanism yet, and I think my proof above is a more readable. *) 