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Proof step 10 from proof inv-inverts (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 11:
n1 = 0, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , leninv(nσ) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[nσ[n]] = n
Proof step 14:
n1 = n2 +1, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , leninv(nσ) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[nσ[n]] = n

Rule argument

var: n1

Additional information

Used simplifier rules: nat freely generated by 0, +1;;