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Proof step 146 from proof inv-get-inverse-between (previous step)

# nσ ≠ ∞, nσ[(# nσ).n - 1] = 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 < 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 147:
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[(# nσ).n - 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[(# nσ).n - 1] = 0, # nσ = ∞

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1