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Proof step 43 from proof propagate (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
H0[r] = H3[r], !(id, O2, P1)(H3), !(me, O2, P)(H0), P0(H), r ∈ H0, r ∈ H0 ∪ H1, r ∈ H3, H3 ⊥ H2, H0 ⊥ H, H1 ⊥ H0, ∀ r0. (O1(r0) = me ↔ O2(r0) = me) ∧ (O1(r0) = me → (r0 ∈ H0 ∨ r0 ∈ H ↔ r0 ∈ H3 ∨ r0 ∈ H2) ∧ (r0 ∈ H0 ∨ r0 ∈ H → (r0 ∈ H0 ⊃ H0[r0];H[r0]) = (H3 ∪ H2)[r0])) ⊦ id = me

Rule argument

Additional information

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