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Proof step 15 from proof finz-bounded (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ odd(n) ↔ (∃ m. n = m + m + 1)
Proof step 14:
∃ m0. lead(m) = m0 + m0 + 1, (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: nat-even instance-name: lemma-name: oex antecedent: succedent: odd(n) ↔ (∃ m. n = m + m + 1) substlist: [n] → [lead(m)] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 0)) dummy: () ]

Additional information

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