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Proof step 38 from proof in-mod (previous step)

sdt0[str][p0] = dnode(md0, sdt), str0 = str, p0 ∈ sdt0[str], str ∈ sdt0, str0 ∈ sdt0[str, sdt0[str][p0, dnode(md1, sdt)]], ∀ md2, sdt1, md3, sdt2, q, md4. dnode(md2, sdt1)[p0] = dnode(md3, sdt2) ∧ p0 ∈ dnode(md2, sdt1) → (q ∈ dnode(md2, sdt1)[p0, dnode(md4, sdt2)] ↔ q ∈ dnode(md2, sdt1)) ⊦ q0 ∈ sdt0[str, sdt0[str][p0, dnode(md1, sdt)]][str0] ↔ str0 ∈ sdt0 ∧ q0 ∈ sdt0[str0], p0 = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
sdt0[str][p0] = dnode(md0, sdt), str ∈ sdt0, p0 ∈ sdt0[str], ∀ md2, sdt1, md3, sdt2, q, md4. dnode(md2, sdt1)[p0] = dnode(md3, sdt2) ∧ p0 ∈ dnode(md2, sdt1) → (q ∈ dnode(md2, sdt1)[p0, dnode(md4, sdt2)] ↔ q ∈ dnode(md2, sdt1)) ⊦ q0 ∈ sdt0[str][p0, dnode(md1, sdt)] ↔ q0 ∈ sdt0[str], p0 = ε

Rule argument

Additional information

Used simplifier rules: ⊦ str ∈ sdt[str, tree]; (s)% ⊦ sdt[str, tree][str] = tree; (s)