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Proof step 54 from proof at-put (previous step)

sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, q0 ∈ dnode(md0, sdt0[str0, dnode(md, sdt[str, tree0])])[str1 '], 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])) ⊦ dnode(md0, sdt0[str0, dnode(md, sdt[str, tree0])])[str1 '][q0] = (str1 ' + q0 = str0 ' ⊃ dnode(md0, sdt0)[str1 ' + q0][restn(# q0, ε) + str ', tree0];dnode(md0, sdt0)[str1 ' + q0]), str0 ' + str ' ⊑ str1 ' + q0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];%nil-at: ⊦ tree[ε] = tree;%nil-in: ⊦ ε ∈ tree;%put-here: ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]);% ⊦ str ' = str0 ' + p ↔ str = str0 ∧ p = ε; (ws)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)% ⊦ str = str0 → sdt[str0, tree][str] = tree; (s)% ⊦ str ' + p ⊑ str0 ' + p0 ↔ str = str0 ∧ p ⊑ p0; (ws)% ⊦ p + ε = p; (s)% ⊦ # ε = 0; (s)% ⊦ restn(0, p) = p; (s)% ⊦ ε + p = p; (s)% ⊦ str ∈ sdt[str, tree]; (s)% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ str ' ≠ ε; (s)