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Proof step 40 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H4 ∪ H3, P(H4), P0(H3), r0 ∈ H4, H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, ∀ r1. ((r = r1 ⊃ O1(r0);O1(r1)) = O1(r0) ↔ O2(r1) = O1(r0)) ∧ ((r = r1 ⊃ O1(r0);O1(r1)) = O1(r0) → (r1 = r ∨ r1 ∈ H1 ↔ r1 ∈ H2) ∧ (r1 = r ∨ r1 ∈ H1 → (H1 ++ r)[r1] = H2[r1])) ⊦ O2(r0) = O1(r0), r = null, r ∈ H1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: r0

Additional information

Used terms: r0