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Proof step 39 from proof propagate (previous step)

(r ∈ H0 ⊃ H0[r];H1[r]) = (r ∈ H3 ⊃ H3[r];H2[r]), r ∈ H0 ∨ r ∈ H1 ↔ r ∈ H3 ∨ r ∈ H2, !(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 ∈ H3 → r ∈ H1) ∧ (r ∈ H3 → H3[r] = H1[r])

User applied rule case distinction

And obtained the following new goals

Proof step 40:
r ∈ H0 ∨ r ∈ H1 ↔ r ∈ H3 ∨ r ∈ H2, H0[r] = (r ∈ H3 ⊃ H3[r];H2[r]), !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), r ∈ H0 ∪ H1, r ∈ H0, H1 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, (r ∈ H3 → r ∈ H1) ∧ (r ∈ H3 → H3[r] = H1[r])
Proof step 45:
r ∈ H0 ∨ r ∈ H1 ↔ r ∈ H3 ∨ r ∈ H2, H1[r] = (r ∈ H3 ⊃ H3[r];H2[r]), !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), r ∈ H0 ∪ H1, H1 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, r ∈ H0, (r ∈ H3 → r ∈ H1) ∧ (r ∈ H3 → H3[r] = H1[r])

Rule argument

pos: left 1

Proof step 39 from proof propagate (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information