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Proof step 28 from proof inv-get-inverse-between (previous step)

# nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ n0. nσ[n0] ≠ n ⊦ nσ[∞.n + m + 1] = n + 1

User applied rule all left

And obtained the following new goals

Proof step 29:
# nσ = ∞, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ n0. nσ[n0] ≠ n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∃ m. nσ[m] = n ⊦ nσ[∞.n + m + 1] = n + 1

Rule argument

pos: left 7 terms: n abort: F discard: T computed: F

Additional information

Used terms: n