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Proof step 5 from proof restN-dups (previous step)

dups(restn(n, a ' + x0)), n ≤ 1 + # x0, ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < n → restn(n, x) = restn(n -1, x.tail)
Proof step 7:
0 < n, dups(restn(n -1, (a ' + x0).tail)), n ≤ 1 + # x0, ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ 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 < m ↔ m ≠ 0; (s)% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ 0 ≤ m; (s) Used terms: a ' + x0, n