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Proof step 3 from proof diff-del (previous step)

⊦ a0 ∈ s1 \ s2 ↔ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
⊦ a0 ∈ s1 \ s2 ↔ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2, a ∈ s1

Rule argument

fma: a ∈ s1

Additional information

Used simplifier rules: diff-del-minus: ⊦ (s -- a) \ (s1 -- a) = (s -- a) \ s1;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%diff-del-minus-3: ⊦ a ∈ s1 → (s -- a) \ s1 = s \ s1; Used formulas: a ∈ s1