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Proof step 40 from proof inv-ls-succ (previous step)

# nσ = ∞, nσ[n] = n2 + 1, n2 + 1 < nσ[m], n < m, stutf(nσ), ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → # nσ = ∞ ∧ stutf(nσ) ∧ (∀ n1. nσ[n1] ≠ n0 + 1) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 41:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 42:
# nσ = ∞, nσ[n] = n2 + 1, n < m, n2 + 1 < nσ[m], stutf(nσ), stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ 2 + n2 ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = 2 + n2), ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → # nσ = ∞ ∧ stutf(nσ) ∧ (∀ n1. nσ[n1] ≠ n0 + 1) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

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, m, 2 + n2]

Additional information

Used terms: nσ, m, 2 + n2