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Proof step 54 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, nσ[0] = 0, ⌜nσ[n1]⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ[n1 + 1] = nσ[n1], nσ[n1 + 1] = nσ[n1] + 1, nσ = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
# nσ = ∞, nσ[0] = 0, ⌜nσ[n1]⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ n. nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ[n1 + 1] = nσ[n1], nσ[n1 + 1] = nσ[n1] + 1, nσ = []

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)