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Proof step 20 from proof propagate (previous step)

H0 ∪ H1 = H3 ∪ H2, !(id, O2, P1)(H3), !(me, O2, P)(H0), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H3 ⊥ H2, H0 ⊥ H, H1 ⊥ H0, ∃ H. H3 ∪ H = H1 ∧ H ⊥ H3 ⊦ id = me, ((!(id, O2, P1) * true))(H1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
H0 ∪ H3 ∪ H4 = H3 ∪ H2, !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H4 ⊥ H3, H3 ∪ H4 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, ((!(id, O2, P1) * true))(H3 ∪ H4)

Proof step 20 from proof propagate (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information