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Proof step 4 from proof dnode-onein (previous step)

tree = dnode(md0, sdt0), str ' ∈ tree ⊦ ∃ md, sdt. tree = dnode(md, sdt) ∧ str ∈ sdt

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt
Proof step 6:
str ∈ sdt0 ⊦ ∃ md, sdt. dnode(md0, sdt0) = dnode(md, sdt) ∧ str ∈ sdt

Rule argument

Additional information

Used simplifier rules: one-ind: ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt;