Additional file 5 Proof S3- Proof of analytical bound 3 2v s t u Theorem 3: z eq hm 1 v 1 u 1 s t Lemma 3.1: x eq u 2 Proof of Lemma 3.1: z 1. at equilibrium 1 v 1 u 1 s t z W hm x v W x 1 W 2. If z=0 then the theorem is self evident 3. 4. z 0 W 2 1 v 1 u 1 s t W hm x v W x W 2 1 v 1 u 1 s t W 1 v 1 u 1 s t hm x v W x W 2 1 v 1 u 1 s t W 5. 1 v 1 u 1 s t v W x 1 v 1 u 1 s t hmx x 6. 7. W 2 1 v 1 u 1 s t W 1 v 1 u 1 s t hm 1 v 1 u 1 s t v W x 1 v 1 u 1 s t hm x W 2 1 v 1 u 1 s t W v 1 v 1 u 1 s t hm 1 8. 9. x W 1 v 1 u 1 s t 1 v 1 u hm Because W 1 s t v W v 1 1 v 1 u W 1 s t x 1 s t 1 v 1 u hm v u v uv W 1 s t 10. x 1 s t 1 v 1 u hm 11. It is biologically plausible that s t 0.25, 3v u 1 s t 2 12. 1 1 W 1 s t 2u 2u 2 u W 1 s t 1 2 u 1 u 13. x 2 9 1 W 1 s t 1 v 1 v uv v W 1 s t W 1 s t 1 s t u v 1 v 1 u hm 14. x 1 s t u v 15. x u (using the above assumptions s t 0.25,3v u ) QED Lemma 3.1 2 Proof of theorem 3: x v W x hm 1 v 1 u 1 s t z 1. At equilibrium x 1 W W 2 xW 2 x v W x 2. 3. hm 1 v 1 u 1 s t z W 1 W hm 1 v 1 u 1 s t z x v W x x v W x W xW 2 x v W x xW hm 1 v 1 u 1 s t x v W x 4. z W 5. x v W x xW z W hm 1 v 1 u 1 s t x z 6. hm 1 v 1 u 1 s t Wv W x hm 1 v 1 u 1 s t x z 7. W 1 W W 1 W hm 1 v 1 u 1 s t v hm 1 v 1 u 1 s t x 8. We notice that 9. W 1 W 1 W sy s t z s y z tz s t v st x z hm 1 v 1 u 1 s t 2v st u 10. z eq using Lemma 3.1 QED hm 1 v 1 u 1 s t Supplementary proof 3a: 3 Theorem 3.a: hm 1. x st 1 v 1 u x vW x z 1 hm 1 v 1 u 1 s t W W x vW x 2. x vW x W z 1 hm 1 v 1 u 1 s t W u Lemma 3.1 0 2 3. W 2 W hm 1 v 1 u 1 s t z x 4. hm 1 v 1 u 1 s t z W W 2 5. hm W 1 W 1 v 1 u 1 s t z sy s t z 6. hm 1 sy s t z 1 v 1 u 1 s t z 7. hm 8. hm 1 s t s t z 1 v 1 u 1 s t z st QED 1 v 1 u Supplementary proof 3b: Theorem 3b: az eq y eq , z eq b 2v hm a 1 s t bu 1 v 1 u 1 s t 4 1. From supplementary proof 3, paragraph 8 in the proof of the main theorem: z W 1 W hm 1 v 1 u 1 s t v hm 1 v 1 u 1 s t x 2v u 2. Using lemma 3.1 we get z hm 1 v 1 u 1 s t 1 W 2v u hm zhm 1 v 1 u 1 s t sy s t z 3. az eq y eq , z eq b 2v 4. hm a 1 s t bu 1 v 1 u 1 s t QED 4v u In particular z eq 0.5 hm 1 v 1 u 1 s t 2s t v s 2 u t z eq In addition, z 0 hm eq 1 v 1 u 1 s t 5
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