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Proof step 7 from proof prefixn-toset-restr-empty (previous step)
| ax.toSet | m = ∅, n ≤ # ax ⊦ #(firstn(n, ax).toSet | m) = 0, firstn(n, ax).toSet | m = ∅ |
Heuristic elimination applied rule insert elim lemma
And closed the goal!
Rule argument
spec: list-del, inst: address-list-set
name: elim-firstn-restn
seq: ⊦ n ≤ # ax
→ ( ax0 = firstn(n, ax) ∧ ax1 = restn(n, ax)
↔ ax = ax0 + ax1 ∧ # ax0 = n)
subst: [ax, n, ax0, ax1] → [ax, n, ax0, ax1]Additional information
Used simplifier rules:
⊦ (ax + ax0).toSet = ax.toSet ∪ ax0.toSet; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ (as0 ∪ as1) | n = (as0 | n) ∪ (as1 | n); (s)%una: ⊦ (as ∪ as0) ∪ as1 = as ∪ as0 ∪ as1; from specification set-union[node-store]%unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[node-store]% ⊦ as ∪ as0 = ∅ ↔ as = ∅ ∧ as0 = ∅; (ws)%⊦ n ≤ # ax
→ ( ax0 = firstn(n, ax) ∧ ax1 = restn(n, ax)
↔ ax = ax0 + ax1 ∧ # ax0 = n); (elim)
Used terms: ax, n, ax0, ax1