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Proof step 59 from proof eq-leninv (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 60:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 61:
# nσ = ∞, nσ[m0] + 1 = nσ[n], nσ[m0 + 1] = nσ[n], m0 < n, nσ[m0] < nσ[n], m0 < m, stutf(nσ), mon≤(nσ) ∧ m0 + 1 ≤ m ∧ ⌜m⌝ < # nσ → nσ[m0 + 1] ≤ nσ[m], ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ nσ[m] = nσ[n]

Rule argument

spec: <toplevel> name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m0 + 1, m]

Additional information

Used terms: nσ, m0 + 1, m