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Proof step 95 from proof inv-get-inverse-between (previous step)

stutf(nσ) ∧ ⌜m1⌝ < # nσ ∧ n ≤ nσ[m1] → (∃ m0. m0 ≤ m1 ∧ nσ[m0] = n), (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n), # nσ = ∞, stutf(nσ), n < nσ[m1], ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[m1] = 0, nσ[∞.n + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1).n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 96:
# nσ = ∞, m0 ≤ m1, stutf(nσ), nσ[m0] < nσ[m1], ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0]) ∨ (∀ n0. nσ[n0] ≠ nσ[m0]), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[m1] = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%lels-03: ⊦ m < n → m ≤ n; from specification nat