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Proof step 5 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ to leninv(nσ).n)[nσ[n]] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
nσ[n + 1] = nσ[n] + 1, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ ) [nσ[n] ] = n

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nnf⌝ to n = ⌜nnf to n⌝ ; (s)