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Proof step 56 from proof leninv-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 57:
# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m] + 1, stutf(nσ), m1 = (m1 + 1) + m - m + 1 ∧ # nσ = ∞ ∧ nσ[(m1 + 1) + m] = nσ[m + 1] + 1 ∧ m + 1 < (m1 + 1) + m ∧ stutf(nσ) → (∃ m2. m + 1 ≤ m2 ∧ m2 < (m1 + 1) + m ∧ nσ[m + 1] = nσ[m2] ∧ nσ[m2 + 1] = nσ[m2] + 1), nσ[m1 + m] = nσ[m] + 1 ∧ m1 ≠ 0 → m ≤ m0 ∧ m0 < m1 + m ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1 ⊦ ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

pos: left 4 terms: (m1 + 1) + m, m + 1, nσ abort: F discard: T computed: F

Additional information

Used terms: (m1 + 1) + m, m + 1, nσ