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Proof step 49 from proof inv-get-inverse-last (previous step)

# nσ = ∞, nσ[m1] = nσ[n1], ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ⌜n1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
# nσ = ∞, nσ[m1] = nσ[n1], stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[n1] + 1, ∀ m0. nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

Additional information

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