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Proof step 65 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, nσ[m] = n1, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜1⌝ < # nσ, stutf(nσ), n ≤ n1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
# nσ = ∞, stutf(nσ), n ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[max∞(λ n0. nσ[n0] = n).n + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: 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% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)% ⊦ ⌜m⌝ < ∞; (s)