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Proof step 2 from proof add-inter-rem (previous step)

⊦ ((s0 ++ a) ∩ s1) -- a = (s0 ∩ s1) -- a, (s0 ++ a) ∩ s1 = s0 ∩ s1

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: a ∈ (s0 ++ a) ∩ s1

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;%int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%intersection-insert: ⊦ ¬ a ∈ s → s ∩ (s0 ++ a) = s ∩ s0;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic Used formulas: a ∈ (s0 ++ a) ∩ s1