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Proof step 16 from proof finz-bigger (previous step)

2 * m0 + 2 = 2 * 2^ m2 + 4 * 2^ m2 * m, #fz(m0) < m2, ∀ m1, n, n0. m1 < m0 + m0 + 1 → (m1 + 1 = n * 2^ n0 ∧ #fz(m1) < n0 → false) ⊦ 2 * (m0 + 1) = 2 * 2^ m2 * (1 + 2 * m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
2 * m0 + 2 = 2 * 2^ m2 + 4 * 2^ m2 * m, #fz(m0) < m2, ∀ m1, n, n0. m1 < m0 + m0 + 1 → (m1 + 1 = n * 2^ n0 ∧ #fz(m1) < n0 → false) ⊦ 2 * m0 + 2 = (4 * m + 2) * 2^ m2

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m * (n + 1) = m * n + m])]]