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Proof step 86 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 87:
nσ[0] = 0, nσ[m + 1] = nσ[m] + 1, # nσ = ∞, nσ[n1] ≤ nσ[m], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ n. nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ = [], nσ[n1 + 1] = nσ[n1] + 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%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf