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Proof step 22 from proof elim-fromto (previous step)

⊦ ( # x = #((λ n. (n < # x ⊃ x[n];nf0(n - # x))) to # x) ∧ (∀ n. n < # x → x[n] = ((λ n. (n < # x ⊃ x[n];nf0(n - # x))) to # x)[n])) ∧ nf0 = (λ n. (n < # x ⊃ x[n];nf0(n - # x))) from # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ nf0 = (λ n. (n < # x ⊃ x[n];nf0(n - # x))) from # x

Rule argument

Additional information

Used simplifier rules: size-inf-to: ⊦ #(nf to n) = n;%get-infto: m < n ⊦ (nf to n)[m] = nf(m);% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)