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Proof step 109 from proof inv-get-inverse-succ (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 110:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 111:
# nσ = ⌜m⌝ + ⌜1⌝ , nσ[n1] = n, n1 < m, stutf(nσ), n ≤ nσ[m], ∀ m0. ¬ m < m0 ∧ nσ[m0] = n → ¬ n1 < m0 ⊦ nσ[n1 + 1] = n + 1, m < n1

Rule argument

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

Additional information

Used terms: n1, λ n0. ¬ m < n0 ∧ nσ[n0] = n