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Proof step 14 from proof split-old (previous step)

⊦ (↓ (P0 * P1))(H, H0) ↔ (↓ P0 * ↓ P1)(H, H0)

User applied rule case distinction

And obtained the following new goals

Proof step 10:
(↓ (P0 * P1))(H, H0) ⊦ (↓ P0 * ↓ P1)(H, H0)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]