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Proof step 13 from proof finz-bounded (previous step)

(m + m0) + n0 = 2^ #fz(m), m + 1 = ((m + m0) + n0) * lead(m), m + m0 = 2^ n, lead(m) = 2 * m1 + 1, n < #fz(m) ⊦ m0 = 0

User applied rule insert equation

And obtained the following new goals

Proof step 12:
(m + m0) + n0 = 2^ #fz(m), m + 1 = ((m + m0) + n0) * (2 * m1 + 1), m + m0 = 2^ n, lead(m) = 2 * m1 + 1, n < #fz(m) ⊦ m0 = 0

Rule argument

[inserteqarg: pos: left 4,F,F,()]

Additional information

[[CONCRETESIMPRULES : ()]]