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Proof step 84 from proof inv-get-inverse-between (previous step)

leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , stutf(nσ), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, n < n1 ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1, n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 85:
⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1);nσ[(# nσ).n - 1] = n))
Proof step 86:
stutf(nσ), (# nσ = ∞ ⊃ ∃ m. nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = n1 + 1);nσ[(# nσ).n - 1] = n1), stutf(nσ), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, n < n1 ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1, n1 = 0

Rule argument

spec: inst: name: leninv-succ seq: ⊦ stutf(nσ) → ( leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1) ;nσ[(# nσ).n - 1] = n )) subst: [n, nσ] → [n1, nσ] flags: rewrite, current

Additional information

Used terms: n1, nσ