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Proof step 30 from proof inv-inverts (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 31:
⊦ 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 < n → m ≤ n; (s)% ⊦ ⌜m⌝ < ∞; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%m < n ⊦ (nnf to n)[m] = nnf(m); (s)% ⊦ ⌜m⌝ .n = m; (s) Used terms: nσ, n, m