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Proof step 50 from proof inv-inverts-leq (previous step)

# nσ = ∞, nσ[n0] = nσ[n], nσ[n1] = nσ[n], m + 1 < n0, ∞.n < n, m + 1 < n1, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 51:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 52:
# nσ = ∞, nσ[n1] = nσ[n], nσ[n0] = nσ[n], m + 1 < n1, ∞.n < n, m + 1 < n0, stutf(nσ), nσ[n] ≤ nσ[m], mon≤(nσ) ∧ m + 1 ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m + 1] ≤ nσ[n0], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

spec: <toplevel> name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m + 1, n0]

Additional information

Used terms: nσ, m + 1, n0