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Proof step 13 from proof subset-is-union (previous step)

(s -- a) ∩ s0 = ∅, a ∈ s ⊦ #(s ∩ s0) = 0, s ∩ s0 = ∅, # s = 0, s = ∅, a ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a
Proof step 15:
(s0 ∩ s) -- a = ∅, a ∈ s ⊦ #(s ∩ s0) = 0, s ∩ s0 = ∅, # s = 0, s = ∅, a ∈ s0

Rule argument

spec: inst: name: int-del-rew seq: ⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a subst: [s, a, s0] → [s0, a, s] flags: rewrite, current

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s; Used terms: s0, a, s