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Proof step 2 from proof diff-in (previous step)

⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

User applied rule induction

And obtained the following new goals

Proof step 3:
∀ H, r, H0. # H < # H1 → true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

Rule argument

indvar: # H1 wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # H1