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Proof step 22 from proof inv-get-inverse-between (previous step)

nσ[n] = nσ[m0], # nσ = ∞, m0 < n, m1 < n, m0 < m1, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, stutf(nσ), ⌜nσ[m0]⌝ + ⌜1⌝ < leninv(nσ), nσ[m0] < nσ[m1], m0 + 1 < n, ∀ m. ∃ m0. m < nσ[m0] ⊦ n = 0, nσ[m1] = 0, m1 = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 23:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 24:
mon≤(nσ) ∧ m1 ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m1] ≤ nσ[m0], nσ[n] = nσ[m0], # nσ = ∞, m0 < n, m1 < n, m0 < m1, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, stutf(nσ), ⌜nσ[m0]⌝ + ⌜1⌝ < leninv(nσ), nσ[m0] < nσ[m1], m0 + 1 < n, ∀ m. ∃ m0. m < nσ[m0] ⊦ n = 0, nσ[m1] = 0, m1 = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0

Rule argument

spec: inst: name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m1, m0] flags: current, []

Additional information

Used terms: nσ, m1, m0