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Proof step 29 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m0] = nσ[m] + 1, m < m0, stutf(nσ), ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ m0 = 0, ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

User applied rule cut formula

And obtained the following new goals

Proof step 30:
∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1, # nσ = ∞, nσ[m0] = nσ[m] + 1, m < m0, stutf(nσ), ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ m0 = 0, ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)
Proof step 49:
# nσ = ∞, nσ[m0] = nσ[m] + 1, m < m0, stutf(nσ), ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ ∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1, m0 = 0, ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Rule argument

fma: ∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1

Additional information

Used formulas: ∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1