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Proof step 10 from proof finz-bounded-03 (previous step)

m0 < 2^ n, ∀ m1, n. m1 < m0 + m0 + 1 → (m1 < 2 * 2^ n → (n = #fz(m1) ↔ m1 + 1 = 2^ n)) ⊦ n = #fz(m0) + 1 ↔ 2 * m0 + 2 = 2^ n

User applied rule constructor cut

And obtained the following new goals

Proof step 9:
n = m +1, m0 < 2^ n, ∀ m1, n. m1 < m0 + m0 + 1 → (m1 < 2 * 2^ n → (n = #fz(m1) ↔ m1 + 1 = 2^ n)) ⊦ n = #fz(m0) + 1 ↔ 2 * m0 + 2 = 2^ n

Rule argument

[var-arg: n]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : nat generated by 0, +1;], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ 2^ 0 = 1], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))])]]