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Proof step 9 from proof inv-inverts-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ∞, max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ∞, stutf(nσ) ⊦ ∞.n = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf