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Proof step 26 from proof diff-in (previous step)

true, ∀ H, r, H0. # H < # H1 → (true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H)) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
∀ H, r, H0. # H < # H1 → (true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H)) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]