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Proof step 18 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m] = n2 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1);nσ[(# nσ).n - 1] = n2) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] = n

User applied rule case distinction

And obtained the following new goals

Proof step 19:
# nσ = ∞, nσ[m] = n2 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1), nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] = n
Proof step 31:
# nσ ≠ ∞, nσ[(# nσ).n - 1] = n2, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] = n

Rule argument

pos: left 5

Proof step 18 from proof inv-inverts (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information