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If (|y|ge R_{3}/2), then by a similar approach we deduce again relation (16).
If (|y|gefrac{R_{2}}{2}), then in a similar way as above, we deduce again relation (14).
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If a − 1 ≠ 0, then N ¯ ( r, 1 G + a − 1 ) = N ¯ ( r, 1 H ). Similar to the discussion in Subcase 2.1, we can deduce a contradiction again.
If a + 1 ≠ 0, then a + 1 is a Picard exceptional value of G. Similar to the discussion in Subcase 2.1, we can deduce a contradiction again.
If a + 1 ≠ 0, then a + 1 is a Picard exceptional value of G. Similar to the discussion as in Subcase 2.1, we can deduce a contradiction again.
If a − 1 ≠ 0, then N ¯ ( r, 1 G + a − 1 ) = N ¯ ( r, 1 F ). Similar to the discussion as in Subcase 2.1, we can deduce a contradiction again.
Applying Lemma 2.2 again, we deduce that (1-(2+C)+2C=0), which implies that (C=1).
Again, we deduce that there exists a constant (K_{1.1}>0) with sup_{x geq0} F_{1}(x) leq K_{1.1}.
Again, we deduce from Lemmas 2.4, 2.5 and 2.6 that S N and N are continuous and completely continuous.
Again, we deduce from Lemmas 2.4, 2.5 and 2.6 that S P and P are continuous and completely continuous.
Again, we deduce that there exists a constant (K_{1.1}>0) with sup_{x geq0} B_{1}(x) leq K_{1.1}.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com