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Proof step 21 from proof in-mod (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 23:
sdt0[str] = dnode(md0, sdt), str ∈ sdt0, str0 ∈ sdt0[str, dnode(md1, sdt)] ⊦ str0 ∈ sdt0[str, dnode(md1, sdt)] ∧ q0 ∈ sdt0[str, dnode(md1, sdt)][str0] ↔ str0 ∈ sdt0 ∧ q0 ∈ sdt0[str0]

Rule argument

spec: <toplevel> name: one-at seq: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str] subst: [str, sdt, md] → [str0, sdt0[str, dnode(md1, sdt)], md] flags: <all paths>

Additional information

Used simplifier rules: ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws) Used terms: str0, sdt0[str, dnode(md1, sdt)], md