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Proof step 2 from proof lslen-fromnotempty-list (previous step)

x from n = [], n < # x ⊦ x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n])

Rule argument

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

Additional information

Used simplifier rules: length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%size-01: ⊦ # [] = 0; from specification list-data%zero-01: ⊦ n < m → m - n ≠ 0; from specification nat Used terms: [], x from n