popper-logic-scientific-discovery

A3
B1
B2
C
p(a, a) = p(b, b).
p(ab, c) � p(a, c),
p(ab, c) = p(a, bc)p(b, c).
p(a, a) ≠ p(b, a) → p(a, a) = p(c, a) + p(c¯, a).
I now proceed to the derivations.
appendix *v 357
(1) p(a, a) = p(b, b) = k Abbreviation based upon A3
(2) p((aa)a, a) � p(aa, a) � p(a, a) = k B1, 1
(3) p((aa)a, a) = p(aa, aa)p(a, a) = k2 B2, 1
(4) k2 � k 2, 3
(5) 0 � k � 1
(6) k ≠ p(a, b) → k = k + p(b¯, b)
(7) k ≠ p(a, b) → p(b¯, b) = 0
4 (and Postulate 2)
C, 1
6
(8) p(ab¯, b) = p(a, b¯b)p(b¯, b)
(9) k ≠ p(a, b) → 0 = p(ab¯, b) � p(a, b)
B2
7, 8, B1
(10) k ≠ p(a, b) → 0 � p(a, b) 9
(11) 0 > p(a, b) → k = p(a, b) 10
(12) k = p(a, b) → 0 � p(a, b) 5
(13) 0 > p(a, b) → 0 � p(a, b) 11, 12
(14) 0 � p(a, b) 13 (or 10, 12)
(15) 0 � p(ā, b) 14
(16) k ≠ p(a, b) → k � p(a, b) C, 1, 15
(17) p(a, b) � k � 1 16, 5
(18) 0 � p(a, b) � k � 1 14, 17
(19) k = p(aa, aa) � p(a, aa) � k 1, B1, 17
(20) k = p(a(aa), a(aa)) � p(a, a(aa)) � k 1, B1, 17
(21) k = p(aa, aa) = p(a, a(aa))p(a, aa) = k2 1, B2, 19, 20
(22) k = k2 21
(23) (Ea) (Eb) p(a, b) ≠ 0 → k = 1 18, 22
(24) (Ea) (Eb) p(a, b) ≠ 0 A1
(25) p(a, a) = k = 1 1, 23, 24