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Proof step 18 from proof finz-bounded (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 17:
2^ n + n0 = 2^ #fz(m), m + 1 = (2^ n + n0) * lead(m), n < #fz(m), ∃ m0. m + m0 = 2^ n ∧ m0 ≠ 0 ⊦

Rule argument

[x0lemmaarg: spec-name: nat-basic2 instance-name: lemma-name: addex antecedent: succedent: m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) substlist: [n, m] → [2^ n, m] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 3)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (2^ n, m)]]