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Proof step 94 from proof inv-get-inverse-between (previous step)

# nσ = ∞, n < nσ[m1], ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n) ⊦ nσ[∞.n + m + 1] = n + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 95:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 96:
# nσ = ∞, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, n < nσ[m1], stutf(nσ), stutf(nσ) ∧ ⌜m1⌝ < # nσ ∧ n ≤ nσ[m1] → (∃ m0. m0 ≤ m1 ∧ nσ[m0] = n), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n) ⊦ nσ[∞.n + m + 1] = n + 1

Rule argument

spec: <toplevel> name: stutf-alltaken seq: ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n) subst: [nσ, m, n] → [nσ, m1, n]

Additional information

Used terms: nσ, m1, n