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Proof step 12 from proof diff-in (previous step)

r ∈ (H0 -- r0) \ H ↔ r ∈ H0 -- r0 ∧ ¬ r ∈ H, ∀ H2, r, H0. # H2 < # (H ++ r0)[r0, o] → true → (r ∈ H0 \ H2 ↔ r ∈ H0 ∧ ¬ r ∈ H2) ⊦ r ∈ (H0 -- r0) \ H ↔ r ∈ H0 ∧ r ≠ r0 ∧ ¬ r ∈ H, r0 = null, r0 ∈ H

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
r ∈ (H0 -- r0) \ H ↔ r ≠ r0 ∧ r ∈ H0 ∧ ¬ r ∈ H, ∀ H2, r, H0. # H2 < # (H ++ r0)[r0, o] → true → (r ∈ H0 \ H2 ↔ r ∈ H0 ∧ ¬ r ∈ H2) ⊦ r ∈ (H0 -- r0) \ H ↔ r ∈ H0 ∧ r ≠ r0 ∧ ¬ r ∈ H, r0 = null, r0 ∈ H

Rule argument

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;