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Proof step 148 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 149:
# nσ = ⌜n1⌝ , stutf(nσ), n < nσ[n1 - 1], max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. n0 < n1 ∧ nσ[n0] = n + 1).n ⊦ nσ[n1 - 1] = 0, nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = n + 1).n = 0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1