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Proof step 46 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , stutf(nσ), ⌜n⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 47:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 48:
# nσ = ∞, nσ[n1] = n ∧ (∀ m. nσ[m] = n → ¬ n1 < m), stutf(nσ), ⌜n⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = n + 1

Rule argument

spec: inst: name: maxinf-fin seq: ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m) subst: [n, np] → [n1, λ n0. nσ[n0] = n] flags: rewrite, current

Additional information

Used terms: n1, λ n0. nσ[n0] = n