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Proof step 18 from proof leq-leninv (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 19:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → 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, n, m]

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ ⌜m⌝ < ∞; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: nσ, n, m