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Proof step 10 from proof Sep-eq (previous step)

∀ a. a ∈ s → aP(a) = aP0(a) ⊦ aP ⊗_ s = aP0 ⊗_ s

User applied rule expand right

And obtained the following new goals

Proof step 9:
∀ a. a ∈ s → aP(a) = aP0(a) ⊦ (aP ⊗_ s)(H) ↔ (aP0 ⊗_ s)(H)

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]