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Proof step 3 from proof subseteq-rem-del (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 5:
a ∈ s1, a ∈ s0, s0 ⊆ s1 ⊦ # s1 = 0, # s0 = 0, s0 = ∅, s1 = ∅, ∀ a0. a0 ∈ s1 \ (s0 -- a) ↔ a0 ∈ (s1 \ s0) ++ a

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [(s1 \ s0) ++ a, s1 \ (s0 -- a)] flags: <all paths>

Additional information

Used terms: (s1 \ s0) ++ a, s1 \ (s0 -- a)