Re: collatz proof algorithm reveals e

You make a mistake in line 1. z(1) = 3.
I'll show you the full proof dump for n = 10.
Maybe this will help you understand.
To create such a dump do (prove c3 10 50 'prove)
The proof starts with n = 0 and finds the cycles for n=0 and n=1,
it was designed to handle more than one function and not
for collatz exclusively.

((0 0) (0 0 0 0 0 0 0) 0 () 0 () 0 ())
((1 4) (1 0 1 0 0 0 0) 1 (0) 0 () 0 ())
((1 2) (2 0 1 0 0 0 1) 1 (0) 0 () 1 (4))
((1 1) (3 0 1 0 0 0 2) 1 (0) 0 () 2 (2 4))
((2 1) (4 0 2 0 0 0 2) 2 (0 1) 2 (2 4) 0 ())
((3 10) (5 1 2 0 0 0 2) 2 (0 1) 1 (4) 0 ())
((3 5) (6 1 2 0 0 0 3) 2 (0 1) 1 (4) 1 (10))
((3 16) (7 1 2 0 0 0 4) 2 (0 1) 1 (4) 2 (5 10))
((3 8) (8 1 2 0 0 0 5) 2 (0 1) 1 (4) 3 (5 10 16))
((3 4) (9 1 2 0 0 0 6) 2 (0 1) 1 (4) 4 (5 8 10 16))
((4 2) (10 1 2 0 0 1 6) 2 (0 1) 5 (4 5 8 10 16) 0 ())
((5 16) (11 2 2 0 0 1 6) 2 (0 1) 4 (5 8 10 16) 0 ())
((6 3) (12 3 2 0 0 1 6) 2 (0 1) 3 (8 10 16) 0 ())
((7 22) (13 3 2 1 0 1 6) 2 (0 1) 3 (8 10 16) 0 ())
((7 11) (14 3 2 1 0 1 7) 2 (0 1) 3 (8 10 16) 1 (22))
((7 34) (15 3 2 1 0 1 8) 2 (0 1) 3 (8 10 16) 2 (11 22))
((7 17) (16 3 2 1 0 1 9) 2 (0 1) 3 (8 10 16) 3 (11 22 34))
((7 52) (17 3 2 1 0 1 10) 2 (0 1) 3 (8 10 16) 4 (11 17 22 34))
((7 26) (18 3 2 1 0 1 11) 2 (0 1) 3 (8 10 16) 5 (11 17 22 34 52))
((7 13) (19 3 2 1 0 1 12) 2 (0 1) 3 (8 10 16) 6 (11 17 22 26 34 52))
((7 40) (20 3 2 1 0 1 13) 2 (0 1) 3 (8 10 16) 7 (11 13 17 22 26 34 52))
((7 20) (21 3 2 1 0 1 14) 2 (0 1) 3 (8 10 16) 8 (11 13 17 22 26 34 40
52))
((7 10) (22 3 2 1 0 1 15) 2 (0 1) 3 (8 10 16) 9 (11 13 17 20 22 26 34
40 52))
((8 4) (23 3 2 1 0 2 15) 2 (0 1) 12 (8 10 11 13 16 17 20 22 26 34 40
52) 0 ())
((9 28) (24 4 2 1 0 2 15) 2 (0 1) 11 (10 11 13 16 17 20 22 26 34 40 52)
0 ())
((9 14) (25 4 2 1 0 2 16) 2 (0 1) 11 (10 11 13 16 17 20 22 26 34 40 52)
1 (28))
((9 7) (26 4 2 1 0 2 17) 2 (0 1) 11 (10 11 13 16 17 20 22 26 34 40 52)
2 (14 28))
((10 5) (27 4 2 2 0 2 17) 2 (0 1) 13 (10 11 13 14 16 17 20 22 26 28 34
40 52) 0 ())
((11 34) (28 5 2 2 0 2 17) 2 (0 1) 12 (11 13 14 16 17 20 22 26 28 34 40
52) 0 ())

e.g.
z(0..10) = 28 is the third number on the last line.
z(7) = 23 - 13 = 10.
z(1) = 4 - 1 = 3.

This dump shows tour "flash of insight" is correct. But note that when
proceeding from n to n+1 the prover is called with arguments
(among others) n+1 and collatz(n+1). I do this to maintain the loop
invariants of the algorithm so although this call to collatz may seem
superfluous I count this as 1 not 0.

.

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