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Proof step 21 from proof inv-get-inverse-last (previous step)

# nσ = ∞ ∧ (∀ n1. ¬ n1 < n0 → nσ[n1] = n), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n) ⊦ nσ[∞.n + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1] ⊦ nσ[∞.n + m + 1] = nσ[m1] + 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%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%add-inf-02: ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf