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Proof step 9 from proof finz-bounded-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m0 < 2 * 2^ m, ∀ m1, n. m1 < m0 + m0 + 1 → (m1 < 2 * 2^ n → (n = #fz(m1) ↔ m1 + 1 = 2^ n)) ⊦ m = #fz(m0) ↔ m0 + 1 = 2^ m

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 2^ n ≠ 0], [CSIMPSEQ : ⊦ 2 * m = 0 ↔ m = 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * n + n = m1 * n ↔ ¬ ¬ (m + 1 = m1 ∨ n = 0)], [CSIMPSEQ : ⊦ 2^ (n + 1) = 2 * 2^ n], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ m +1 = m + 1])]]