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Proof step 26 from proof in-mod (previous step)

sdt0[str] = dnode(md0, sdt), str0 = str, str ∈ sdt0, str0 ∈ sdt0[str, dnode(md1, sdt)] ⊦ q0 ∈ sdt0[str, dnode(md1, sdt)][str0] ↔ str0 ∈ sdt0 ∧ q0 ∈ sdt0[str0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
sdt0[str] = dnode(md0, sdt), str ∈ sdt0 ⊦ q0 ∈ dnode(md1, sdt) ↔ q0 ∈ dnode(md0, sdt)

Rule argument

Additional information

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