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Proof step 2 from proof And-unfold (
previous step
)
a ∈ st ⊦ adP ∧_ st = (adP(a, st[a]) ∧ adP ∧_ (st -- a))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
a ∈ st ⊦ adP ∧_ st = (adP(a, st[a]) ∧ adP ∧_ (st -- a))
Rule argument
Additional information