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Proof step 32 from proof modify-agree (previous step)

r ∈ H ↔ r ∈ H1, H = H0 -- r2, r ∈ rs, r2 ∈ H0, r ∈ H1 → H[r] = H1[r] ⊦ # rs = 0, rs = ∅, r2 ∈ rs, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
(H0 -- r2)[r] = H1[r], r2 ∈ H0, r ∈ rs, r ∈ H0, r ∈ H1 ⊦ # rs = 0, H0[r] = H1[r], rs = ∅, r2 ∈ rs

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H; (s)