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Proof step 25 from proof sep-ass (
previous step
)
⊦ ((P0 * P1) * P2)(H) ↔ (P0 * P1 * P2)(H)
User applied rule case distinction
And obtained the following new goals
Proof step 12
:
((P0 * P1) * P2)(H) ⊦ (P0 * P1 * P2)(H)
Rule argument
[CASEDARG : pos: right 1, ()]
Additional information
[[CONCRETESIMPRULES : ()]]