And therefore, if P + Q= zero, Q = zero and equally P = zero. (6) If PQ = zero, P = zero or Q = zero, consequently i. e. Whence, if PQ = zero, the end result follows. (7) If PQ = 1, P = 1 and Q = l, and accordingly P = l; equally Q = l. (8) If a/h = zero, ab/h = Q and a/bh = zero if bh isn't inconsistent. For and because a/h = zero, whence the outcome by means of VI. hence, if a end is most unlikely, we may possibly upload to the realization or upload continuously to the premisses with out affecting the argument. (9) If a/h = 1, a/bh = 1 if bh isn't inconsistent. because a/h = l, by means of (8) if bh isn't inconsistent, whence therefore we may perhaps upload to premisses, which make a end definite, the other premisses no longer inconsistent with them, with out affecting the outcome. (10) If a/h = 1, ab/h = b/ah = b/h, given that a/h = 1, a/bh = 1 through (9) except b/h = zero, whence the end result, except b/h = zero. If b/h = zero, the end result follows from (8). (11) If ab/h = l, a/h = 1. For (12) If and (12. 1) If and hx isn't really inconsistent, and and this is often the primary of equivalence. In advantage of it and of axiom (ii. ), if , we will replacement a for b and vice versa, anywhere they happen in a chance whose premisses comprise h. (13) a/a = 1, except a is inconsistent. For whence a/aa = l via (ii. ), except a/a = zero, i. e a/a = 1, until a is inconsistent via (iii. ), (12), and VI. (13. 1) , until a is inconsistent. This follows from (13) and (1. 1). (13. 2) , until is inconsistent. This follows from (iii. ) by way of writing for a in (13. 1). (14) If a/b = zero and a isn't inconsistent, b/a = zero. permit f be the gang of assumptions, universal to a and b, which we now have imagined to be integrated in each genuine crew; whence therefore, if a is most unlikely given b, then b is most unlikely given a. (15) If h1/h2 = zero, h1h2/h = zero, and because h1/h2 = zero, h1h2/h = zero by way of (8), until h/h2 = zero, whence the end result by means of (iv. b), until h/h2 = zero. If due to the fact that we imagine that h isn't inconsistent, and as a result therefore, if h1 is most unlikely given h2, h1h2 is usually most unlikely and is excluded from each workforce. (15. 1) If h1h2/h = zero and h2h isn't really inconsistent, zero. This, that is the communicate of (15), follows from X. and (6). (16) If h1/h2 = 1, (16. 1) We may possibly write (16): If h1/h2 = 1, (h2 h1)/h = 1, the place ‘’ symbolises ‘implies. ’ hence if h1 follows from h2, then it truly is continually yes that h2 implies h1 (16. 2) If and h2h isn't really inconsistent, h1/h2h = 1. this is often the communicate of (14). (16. three) We may perhaps write (16. 2): If (h2 h1)/h = 1 and h2h isn't inconsistent, h1/h2h = l. hence, if we outline a ‘group’ as a collection of propositions, which persist with from and are yes really to the proposition which specifies them, this proposition proves that, if h2 h1 and h2 belong to a bunch h2h, then h1 additionally belongs to this team. (17) If (h1 :a ≡b)/h = 1 and h1h isn't really inconsistent, a/h1h = b/h1h. This follows from (16. three) and (12). (18) a/a = 1 or . If a is inconsistent, a/h = zero, the place h isn't really inconsistent, and as a result hence except a is inconsistent, isn't inconsistent, and hence (19) In both case by way of (15).