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Proof step 16 from proof propagate (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
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 ⊦ id = me, ((!(id, O2, P1) * true))(H1)

Rule argument

Additional information

Used simplifier rules: ⊦ true(H); (s)