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Proof step 16 from proof finz-bounded (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ odd(lead(n))
Proof step 15:
odd(lead(m)), (m + m0) + n0 = 2^ #fz(m), m + 1 = ((m + m0) + n0) * lead(m), m + m0 = 2^ n, n < #fz(m) ⊦ m0 = 0

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: lead-is-odd antecedent: succedent: odd(lead(n)) substlist: [n] → [m] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (m)]]