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Proof step 92 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 93:
# nσ = ⌜n1⌝ , stutf(nσ), n ≤ nσ[n1 - 1], 1 + max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n < n1 ⊦ nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + 1] = n + 1, n1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic