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Proof step 55 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n2] = nσ[n1] + 1, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ nσ[n1 + m + 1] = nσ[n2]

User applied rule insert lemma

And obtained the following new goals

Proof step 56:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 57:
# nσ = ∞, nσ[n2] = nσ[n1] + 1, n1 + m + 1 < n2, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), mon≤(nσ) ∧ n2 -1 ≤ n2 ∧ ⌜n2⌝ < # nσ → nσ[n2 -1] ≤ nσ[n2], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n2]

Rule argument

spec: <toplevel> name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, n2 -1, n2]

Additional information

Used terms: nσ, n2 -1, n2