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Proof step 15 from proof elim-fromto (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n])
Proof step 17:
⊦ (# 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

Rule argument

spec: list-del inst: name: extension seq: ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n]) subst: [y, x] → [(λ n. (n < # x ⊃ x[n];nf0(n - # x))) to # x, x] flags: rewrite

Additional information

Used terms: (λ n. (n < # x ⊃ x[n];nf0(n - # x))) to # x, x