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Proof step 84 from proof inv-get-inverse-between (previous step)

leninv(nσ) = ⌜n0⌝ , n0 = n1 +1, 1 + n < n0, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, stutf(nσ) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , n < n1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, stutf(nσ) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ m +1 = m + 1; (s)