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Proof step 5 from proof subseteq-rem-del (previous step)

a ∈ s1, a ∈ s0, s0 ⊆ s1 ⊦ # s1 = 0, # s0 = 0, s0 = ∅, s1 = ∅, ∀ a0. a0 ∈ s1 \ (s0 -- a) ↔ a0 ∈ (s1 \ s0) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s0, a ∈ s1, s0 ⊆ s1 ⊦ # s0 = 0, # s1 = 0, a0 ∈ s1 \ (s0 -- a) ↔ a0 = a ∨ a0 ∈ s1 \ s0, s0 = ∅, s1 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)