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Proof step 5 from proof inv-get-inverse-succ (previous step)

⌜n⌝ < leninv(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)[n] ⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ 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)[n] + 1 ] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⌜n⌝ < leninv(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⌝ ) [n ] ⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ 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⌝ )[n] + 1 ] = n + 1

Rule argument

Additional information

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