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Proof step 83 from proof inv-get-inverse-between (previous step)

n0 = n1 +1, leninv(nσ) = ⌜n0⌝ , stutf(nσ), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, 1 + n < n0 ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 84:
leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , stutf(nσ), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, n < n1 ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1, n1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ: ⊦ m +1 = m + 1; from specification nat-basic2