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Proof step 4 from proof subseteq-rem-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ s0, s0 ⊆ s1, a ∈ s1 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, a0 ∈ s1 \ (s0 -- a) ↔ a0 = a ∨ a0 ∈ s1 \ s0

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic