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Proof step 61 from proof inv-get-inverse-last (previous step)

# nσ = ⌜n⌝ , nσ[n - 1] = nσ[n1], ⌜n1⌝ < # nσ, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n1] + 1, # nσ = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
# nσ = ⌜n⌝ , nσ[n - 1] = nσ[n1], n1 < n, 1 + m + n1 < n, stutf(nσ), ∀ m0. m0 < n ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%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%ls-natinf-to-nat-3-1: ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n1 + n + m < n0; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic