Here's a simple derivation of B, C, and I from the pair {CCCCpqqrCCrsCps, Cpp}: |
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1. CCCCpqqrCCrsCps 2. Cpp [I] D1.1 = 3. CCCCpqCrqsCCrps D1.2 = 4. CCCCpqqrCpr D3.2 = 5. CCpqCCqrCpr [B] D3.4 = 6. CCpCqrCqCpr [C] And from here, two additional familiar formulas follow immediately: D4.2 = 7. CpCCpqq [Pon] D7.2 = 8. CCCppqq [Specialized Assertion] |
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