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Proof step 12 from proof from-append-add (previous step)

n ≤ # x0 ⊦ (x0 to n) + x from # x + n = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: append-from-gt seq: ⊦ # x < n → x + y from n = y from (n - # x) subst: [n, y, x] → [# x + n, x, x0 to n] flags: rewrite, current

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;%from-all: ⊦ x from # x = [];%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%eq-01: ⊦ m + n - m = n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: # x + n, x, x0 to n