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Proof step 34 from proof at-put (previous step)

sdt0[str0][p0] = dnode(md, sdt), q = str1 + q0, q ∈ dnode(md0, sdt0)[str0 ' + p0, dnode(md, sdt[str, tree0])], p0 ∈ sdt0[str0], str0 ∈ sdt0, ∀ md0, sdt0, md, sdt, q, str, tree0. dnode(md0, sdt0)[p0] = dnode(md, sdt) ∧ p0 ∈ dnode(md0, sdt0) ∧ q ∈ dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])] → dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])][q] = (p0 + str ' ⊑ q ⊃ tree0[restn(# p0, q.tail)];(q ⊑ p0 ⊃ dnode(md0, sdt0)[q][restn(# q, p0) + str ', tree0];dnode(md0, sdt0)[q])) ⊦ dnode(md0, sdt0)[str0 ' + p0, dnode(md, sdt[str, tree0])][q] = ((str0 ' + p0) + str ' ⊑ q ⊃ tree0[restn(# p0 + 1, q.tail)] ;(q ⊑ str0 ' + p0 ⊃ dnode(md0, sdt0)[q][restn(# q, str0 ' + p0) + str ', tree0];dnode(md0, sdt0)[q]) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: one-restn: p ≠ ε ⊦ restn(n + 1, p) = restn(n, p.tail);%split-in: ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p];%split-at: ⊦ p ∈ tree → tree[p + q] = tree[p][q];% ⊦ restn(n + 1, str ' + p) = restn(n, p); (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(str ' + p) = (# p)+1; (s)%c: ⊦ m + n = n + m; from specification nat-basic[directory-tree-inst]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic[directory-tree-inst]% ⊦ p + str ' ≠ ε; (s)% ⊦ str ' = str0 ' + p ↔ str = str0 ∧ p = ε; (ws)% ⊦ str ' + p ≠ ε; (s)% ⊦ p ⊑ str ' ↔ p = ε ∨ p = str '; (s)% ⊦ p + ε = p; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ (str ' + p).tail = p; (s)% ⊦ str + p = str ' + p; (s)