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Proof step 59 from proof inv-get-inverse-between (previous step)

nσ[m1 + 1] = nσ[n1] + 1, # nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[m1] ≤ nσ[n1] + 1, n1 + m < m1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n1] + 1 → ¬ m1 + 1 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n1] + 1, m1 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 60:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 61:
mon≤(nσ) ∧ n1 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ[n1] ≤ nσ[n1 + m + 1], nσ[m1 + 1] = nσ[n1] + 1, # nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[m1] ≤ nσ[n1] + 1, n1 + m < m1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n1] + 1 → ¬ m1 + 1 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n1] + 1, m1 = 0

Rule argument

spec: inst: name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, n1, n1 + m + 1] flags: current, []

Additional information

Used terms: nσ, n1, n1 + m + 1