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Proof step 22 from proof inv-inverts (previous step)

m < n, nσ[n + 1] = nσ[n] + 1, # nσ = ∞, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , nσ[n] = nσ[m] + 1, m < n, stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[n] ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[n])[nσ[n]] = n, n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1