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Proof step 5 from proof write-again (previous step)

r ∈ H ⊦ ∀ r0. (r0 ∈ H[r, H[r]] ↔ r0 ∈ H) ∧ (r0 ∈ H[r, H[r]] → H[r, H[r]][r0] = H[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
r ∈ H, r0 ∈ H ⊦ (r0 = r ⊃ H[r];H[r0]) = H[r0]

Rule argument

Additional information

Used simplifier rules: in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%at-write: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]);