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Proof step 56 from proof propagate (previous step)

O1(r) = me ↔ O2(r) = me, !(me, O2, P)(H0), P0(H), H0 ⊥ H, O2(r) = me → (r ∈ H0 ∨ r ∈ H ↔ r ∈ H2) ∧ (r ∈ H0 ∨ r ∈ H → (r ∈ H0 ⊃ H0[r];H[r]) = H2[r]), ∀ r0. (O1(r0) = me ↔ O2(r0) = me) ∧ (O1(r0) = me → (r0 ∈ H0 ∨ r0 ∈ H ↔ r0 ∈ H2) ∧ (r0 ∈ H0 ∨ r0 ∈ H → (r0 ∈ H0 ⊃ H0[r0];H[r0]) = H2[r0])) ⊦ id = me, (r ∈ H0 → r ∈ H2) ∧ (r ∈ H0 → H0[r] = H2[r])

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: owned-r: ⊦ !(me, O, P)(H) ∧ O(r) ≠ me → ¬ r ∈ H;%owned-r: ⊦ !(me, O, P)(H) ∧ O(r) ≠ me → ¬ r ∈ H;