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Proof step 21 from proof inv-get-inverse (previous step)

nσ[n] = nσ[n0], # nσ = ∞, n0 < n, m < n, stutf(nσ), n0 < m, nσ[n0] < nσ[m], n0 + 1 < n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = nσ[n0], n0 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 22:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 23:
mon≤(nσ) ∧ m ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m] ≤ nσ[n0], nσ[n] = nσ[n0], # nσ = ∞, n0 < n, m < n, stutf(nσ), n0 < m, nσ[n0] < nσ[m], n0 + 1 < n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = nσ[n0], n0 = 0

Rule argument

spec: inst: name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m, n0] flags: current, []

Additional information

Used terms: nσ, m, n0