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Proof step 19 from proof from-append-add (previous step)

⊦ (nf to n) + x from # x + n = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del name: extension seq: ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n]) subst: [y, x] → [[], (nf to n) + x from # x + n] flags: <all paths>

Additional information

Used simplifier rules: size-inf-to: ⊦ #(nf to n) = n;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n - n + n1 = m - n1; (s)% ⊦ m - m = 0; (s)% ⊦ # [] = 0; (s)% ⊦ ¬ m < 0; (s) Used terms: [], (nf to n) + x from # x + n