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Proof step 7 from proof at-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ tree[ε] = tree
Proof step 9:
⊦ tree[ε, tree0] = tree0
Proof step 10:
⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0])
Proof step 11:
⊦ ε ∈ tree
Proof step 12:
q ∈ dnode(md, sdt[str, tree0]) ⊦ dnode(md, sdt[str, tree0])[q] = (str ' ⊑ q ⊃ tree0[q.tail];(q = ε ⊃ dnode(md, sdt[str, tree0]);dnode(md, sdt)[q]))

Rule argument

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%nil-put: ⊦ tree[ε, tree0] = tree0;%put-here: ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]);%nil-in: ⊦ ε ∈ tree;% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ 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)