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Proof step 46 from proof propagate (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ r ∈ H0 ∧ H0 ⊥ H1 → (r ∈ H1 ↔ false); (s)