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Proof step 20 from proof finz-bounded (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 19:
∃ n0. 2^ n + n0 = 2^ #fz(m), m + 1 = 2^ #fz(m) * lead(m), n < #fz(m), m < 2^ n ⊦

Rule argument

[x0lemmaarg: spec-name: nat instance-name: lemma-name: le-addex antecedent: succedent: m ≤ n ↔ (∃ n0. m + n0 = n) substlist: [n, m] → [2^ #fz(m), 2^ n] 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 : (2^ #fz(m), 2^ n)]]