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Proof step 30 from proof leninv-inf (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 31:
# nσ = ∞, stutf(nσ), ∃ m. nσ[m] = nσ[m1] + 1, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1), ∀ m2. m1 < m2 → nσ[m2] = nσ[m1] ⊦

Rule argument

pos: left 3 terms: nσ[m1] + 1 abort: F discard: F computed: F

Additional information

Used terms: nσ[m1] + 1