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Proof step 2 from proof restn-empty (previous step)

∀ n. n ≤ # x0 → (restn(n, x0) = [] ↔ n = # x0) ⊦ n ≤ #(a + x0) → (restn(n, a + x0) = [] ↔ n = #(a + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
restn(n, a ' + x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = # x0 + 1

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%restN-all: ⊦ restn(# x, x) = [];%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list