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Proof step 65 from proof at-put (previous step)

sdt0[str0] = dnode(md, sdt), str1 = str0, q0 ∈ sdt0[str0, dnode(md, sdt[str, tree0])][str1], str0 ∈ sdt0, str1 ∈ sdt0[str0, dnode(md, sdt[str, tree0])], ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) ∧ q ∈ dnode(md2, sdt2[str2, tree]) → dnode(md2, sdt2[str2, tree])[q] = (str2 ' ⊑ q ⊃ tree[q.tail];(q = ε ⊃ dnode(md1, sdt1[str2, tree]);dnode(md1, sdt1)[q])) ⊦ str1 ' + q0 = str0 ', dnode(md0, sdt0[str0, dnode(md, sdt[str, tree0])])[str1 '][q0] = dnode(md0, sdt0)[str1 ' + q0], str0 ' + str ' ⊑ str1 ' + q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 67:
⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]
Proof step 68:
sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, q0 ∈ dnode(md, sdt[str, tree0]), ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) ∧ q ∈ dnode(md2, sdt2[str2, tree]) → dnode(md2, sdt2[str2, tree])[q] = (str2 ' ⊑ q ⊃ tree[q.tail];(q = ε ⊃ dnode(md1, sdt1[str2, tree]);dnode(md1, sdt1)[q])) ⊦ dnode(md, sdt[str, tree0])[q0] = dnode(md, sdt)[q0], q0 = ε, str ' ⊑ q0

Rule argument

Additional information

Used simplifier rules: one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];%at-dnode: ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p];% ⊦ str ' + p ⊑ str ' + p0 ↔ p ⊑ p0; (s)% ⊦ str ∈ sdt[str, tree]; (s)% ⊦ p = p + p0 ↔ p0 = ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ sdt[str, tree][str] = tree; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)