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Proof step 6 from proof in-put (previous step)

dnode(md0, sdt0)[ε] = dnode(md, sdt), ε ∈ dnode(md0, sdt0) ⊦ q ∈ dnode(md0, sdt0)[ε, dnode(md, sdt[str, tree0])] ↔ (ε + str ' ⊑ q ⊃ restn(# ε, q.tail) ∈ tree0;q ∈ dnode(md0, sdt0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ tree[ε] = tree
Proof step 8:
⊦ tree[ε, tree0] = tree0
Proof step 9:
⊦ ε ∈ tree
Proof step 10:
⊦ q ∈ dnode(md, sdt[str, tree0]) ↔ (str ' ⊑ q ⊃ q.tail ∈ tree0;q ∈ dnode(md, sdt))

Rule argument

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%nil-put: ⊦ tree[ε, tree0] = tree0;%nil-in: ⊦ ε ∈ tree;% ⊦ restn(0, p) = p; (s)% ⊦ # ε = 0; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ ε + p = p; (s)% ⊦ dnode(md, sdt) = dnode(md0, sdt0) ↔ md = md0 ∧ sdt = sdt0; (ws)