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Proof step 100 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n] = nσ[m0], m0 < n, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, nσ[m0] < nσ[m1], m1 < n, stutf(nσ), m0 ≤ m1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[∞.n + m + 1] = nσ[m0] + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 101:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 102:
# nσ = ∞, nσ[n] = nσ[m0], m1 < n, nσ[m0] < nσ[m1], ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, m0 < n, stutf(nσ), m0 ≤ m1, mon≤(nσ) ∧ m1 ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m1] ≤ nσ[m0], ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[∞.n + m + 1] = nσ[m0] + 1

Rule argument

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

Additional information

Used terms: nσ, m1, m0