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Proof step 11 from proof And-discard (previous step)

∀ s1, P. # s1 < #(s0 ++ a) → true → # s1 = 0 ∨ s1 = ∅ ∨ (λ a. P) ∧_ s1 = P ⊦ (P ∧ (λ a. P) ∧_ s0) = P, a ∈ s0

User applied rule expand right

And obtained the following new goals

Proof step 12:
∀ s1, P. # s1 < #(s0 ++ a) → true → # s1 = 0 ∨ s1 = ∅ ∨ (λ a. P) ∧_ s1 = P ⊦ ((P ∧ (λ a. P) ∧_ s0))(H) ↔ P(H), a ∈ s0

Rule argument

pos: right 1

Additional information