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Proof step 80 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n)[n] + m + 1 < ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n)[n + 1], ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦ nσ[((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n)[n] + m + 1] = n + 1, leninv(nσ) = ∞

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜nl⌝ [n] = nl[n]; (s)