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Proof step 25 from proof inv-inverts (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# nσ = ∞, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), n ≤ m, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)