Hence P(A) > (1-a)(1-b) implies P(A) > (1-a-b). Since I let these quantities get arbitrarily close to zero, the quadratic difference term doesn’t matter.
P(A) ≥ (1-a)(1-b) U (a,b)>0 implies P(A) > (1-a-b). That’s a (very slightly) stronger form. I noticed what I thought was an error where you were adding improbabilities together instead of multiplying.
(1-a)(1-b) = 1-a-b+ab > 1-a-b.
Hence P(A) > (1-a)(1-b) implies P(A) > (1-a-b). Since I let these quantities get arbitrarily close to zero, the quadratic difference term doesn’t matter.
P(A) ≥ (1-a)(1-b) U (a,b)>0 implies P(A) > (1-a-b). That’s a (very slightly) stronger form. I noticed what I thought was an error where you were adding improbabilities together instead of multiplying.
The proof doesn’t need the stronger form, however.