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Proof step 35 from proof propagate (previous step)

r ∈ H0 ∪ H1 ↔ r ∈ H3 ∨ r ∈ H2, (H0 ∪ H1)[r] = (r ∈ H3 ⊃ H3[r];H2[r]), !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H1 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, (r ∈ H3 → r ∈ H1) ∧ (r ∈ H3 → H3[r] = H1[r]), r ∈ H0 ∪ H1 ∧ H0 ⊥ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
(H0 ∪ H1)[r] = H3[r], !(id, O2, P1)(H3), !(me, O2, P)(H0), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), r ∈ H3, r ∈ H0 ∪ H1, H3 ⊥ H2, H0 ⊥ H, H1 ⊥ H0 ⊦ id = me, r ∈ H0, r ∈ H1

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)% ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)