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Proof step 22 from proof at-uni (previous step)

(H0 ∪ H)[r] = H[r], r ∈ H, H0 ⊥ H, ∀ H2, H0, r. # H2 < # (H ++ r0)[r0, o] → H0 ⊥ H2 ∧ (r ∈ H0 ∨ r ∈ H2) → (H0 ∪ H2)[r] = (r ∈ H0 ⊃ H0[r];H2[r]) ⊦ H[r] = (H ++ r0)[r], r0 = null, r0 ∈ H, r0 ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
r ∈ H, r0 ≠ null, ¬ r0 ∈ H ⊦ r ≠ r0 → (H ++ r0)[r] = H[r]

Rule argument

spec: <toplevel> name: at-alloc seq: r ∈ H, r0 ≠ null, ¬ r0 ∈ H ⊦ r ≠ r0 → (H ++ r0)[r] = H[r] subst: [r, r0, H] → [r, r0, H] flags: <all paths>

Additional information

Used terms: r, r0, H