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Proof step 88 from proof inv-get-inverse-between (previous step)

n < n1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n1 ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n1) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

User applied rule case distinction

And obtained the following new goals

Proof step 89:
# nσ = ∞, nσ[m1] = n1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, n < n1, stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1
Proof step 147:
nσ[(# nσ).n - 1] = n1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1).n, n < n1, stutf(nσ) ⊦ # nσ = ∞, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

pos: left 4

Proof step 88 from proof inv-get-inverse-between (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information