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Proof step 29 from proof modify-agree (previous step)

r ∈ H ↔ r ∈ H1, H = H0 ++ r0, r ∈ rs, r ∈ H1 → H[r] = H1[r] ⊦ # rs = 0, r0 = null, rs = ∅, r0 ∈ H0, r0 ∈ rs, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)