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Proof step 25 from proof modify-agree (previous step)

r ∈ H ↔ r ∈ H1, modify(rs)(H0, H), r ∈ rs, r ∈ H1 → H[r] = H1[r] ⊦ # rs = 0, rs = ∅, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ modify(rs)(H0, H1) ↔ (∃ r. ¬ r ∈ rs ∧ r ≠ null ∧ ¬ r ∈ H0 ∧ H1 = H0 ++ r) ∨ (∃ r, o. ¬ r ∈ rs ∧ r ∈ H0 ∧ H1 = H0[r, o]) ∨ (∃ r. ¬ r ∈ rs ∧ r ∈ H0 ∧ H1 = H0 -- r)
Proof step 27:
r ∈ H ↔ r ∈ H1, r ∈ rs, r ∈ H1 → H[r] = H1[r], (∃ r. ¬ r ∈ rs ∧ r ≠ null ∧ ¬ r ∈ H0 ∧ H = H0 ++ r) ∨ (∃ r, o. ¬ r ∈ rs ∧ r ∈ H0 ∧ H = H0[r, o]) ∨ (∃ r. ¬ r ∈ rs ∧ r ∈ H0 ∧ H = H0 -- r) ⊦ # rs = 0, rs = ∅, (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Rule argument

spec: <toplevel> name: modify seq: ⊦ modify(rs)(H0, H1) ↔ (∃ r. ¬ r ∈ rs ∧ r ≠ null ∧ ¬ r ∈ H0 ∧ H1 = H0 ++ r) ∨ (∃ r, o. ¬ r ∈ rs ∧ r ∈ H0 ∧ H1 = H0[r, o]) ∨ (∃ r. ¬ r ∈ rs ∧ r ∈ H0 ∧ H1 = H0 -- r) subst: [H1, H0, rs] → [H, H0, rs] flags: <all paths>

Additional information

Used terms: H, H0, rs