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Proof step 186 from proof inv-get-inverse-between (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 187:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 188:
mon≤(nσ) ∧ n1 + m + 1 ≤ n ∧ ⌜n⌝ < # nσ → nσ[n1 + m + 1] ≤ nσ[n], nσ[n] = nσ[n1] + 1, # nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] ≤ nσ[m0], n1 + m + 1 < m0, n ≤ m0, n1 + m + 1 < n, nσ[n1] < nσ[m0], stutf(nσ), n1 < m0, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m ⊦ n = 0, nσ[n1 + m + 1] = nσ[n], nσ[m0] = 0, m0 = 0, n1 = m0

Rule argument

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

Additional information

Used terms: nσ, n1 + m + 1, n