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Proof step 32 from proof inv-get-inverse-last (previous step)

∀ n0. ⌜n0⌝ < # nσ → nσ[n0] < n, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n) ⊦ nσ[∞.n + m + 1] = n + 1

User applied rule case distinction

And obtained the following new goals

Proof step 33:
# nσ = ∞, nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1), ∀ n0. ⌜n0⌝ < # nσ → nσ[n0] < n, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ nσ[∞.n + m + 1] = n + 1
Proof step 35:
# nσ ≠ ∞, nσ[(# nσ).n - 1] = n, ∀ n0. ⌜n0⌝ < # nσ → nσ[n0] < n, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ nσ[∞.n + m + 1] = n + 1

Rule argument

pos: left 5

Proof step 32 from proof inv-get-inverse-last (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information