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Proof step 28 from proof modify-agree (previous step)

r ∈ H ↔ r ∈ H1, r ∈ rs, r ∈ H1 → H[r] = H1[r], ¬ r0 ∈ rs ∧ r0 ≠ null ∧ ¬ r0 ∈ H0 ∧ H = H0 ++ r0 ∨ ¬ r1 ∈ rs ∧ r1 ∈ H0 ∧ H = H0[r1, o] ∨ ¬ r2 ∈ rs ∧ r2 ∈ H0 ∧ H = H0 -- r2 ⊦ # rs = 0, rs = ∅, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

User applied rule case distinction

And obtained the following new goals

Proof step 29:
r ∈ H ↔ r ∈ H1, H = H0 ++ r0, r ∈ rs, r ∈ H1 → H[r] = H1[r] ⊦ # rs = 0, r0 = null, rs = ∅, r0 ∈ H0, r0 ∈ rs, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])
Proof step 32:
r ∈ H ↔ r ∈ H1, H = H0 -- r2, r ∈ rs, r2 ∈ H0, r ∈ H1 → H[r] = H1[r] ⊦ # rs = 0, rs = ∅, r2 ∈ rs, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Rule argument

pos: left 4

Additional information

Used simplifier rules: r0 ∈ H ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)%r ∈ H ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)