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Proof step 4 from proof restn-empty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < n → restn(n, x) = restn(n -1, x.tail)
Proof step 6:
restn(n -1, (a ' + x0).tail) = [], 0 < n, n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = 1 + # x0

Rule argument

spec: <toplevel> name: rest-rec seq: ⊦ 0 < n → restn(n, x) = restn(n -1, x.tail) subst: [x, n] → [a ' + x0, n] flags: <all paths>

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + x0, n