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Proof step 65 from proof inv-inverts-leq (previous step)

# nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < m, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ), n ≤ m, nσ[n] ≤ nσ[m] ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 66:
# nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ∞, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < n, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < m, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m ⊦
Proof step 73:
# nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ⌜n1⌝ , n1 < n, n1 < m, stutf(nσ), nσ[n] ≤ nσ[m], n ≤ m ⊦ max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]) = ∞

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, n] → [max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]), n1]

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]), n1