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Proof step 27 from proof modify-agree (previous step)

r ∈ H ↔ r ∈ H1, r ∈ rs, r ∈ H1 → H[r] = H1[r], (∃ r. ¬ r ∈ rs ∧ r ≠ null ∧ ¬ r ∈ H0 ∧ H = H0 ++ r) ∨ (∃ r, o. ¬ r ∈ rs ∧ r ∈ H0 ∧ H = H0[r, o]) ∨ (∃ r. ¬ r ∈ rs ∧ r ∈ H0 ∧ H = H0 -- r) ⊦ # rs = 0, rs = ∅, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
r ∈ H ↔ r ∈ H1, r ∈ rs, r ∈ H1 → H[r] = H1[r], ¬ r0 ∈ rs ∧ r0 ≠ null ∧ ¬ r0 ∈ H0 ∧ H = H0 ++ r0 ∨ ¬ r1 ∈ rs ∧ r1 ∈ H0 ∧ H = H0[r1, o] ∨ ¬ r2 ∈ rs ∧ r2 ∈ H0 ∧ H = H0 -- r2 ⊦ # rs = 0, rs = ∅, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Proof step 27 from proof modify-agree (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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