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Proof step 7 from proof And-discard (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ s, ∀ s0, P. # s0 < # s → true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P ⊦ # s = 0, (λ a. P) ∧_ s = P

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)