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Proof step 6 from proof toSet-diff-new (previous step)

a ∈ x0 ⊦ a0 ∈ (s0 ++ a) \ x0.toSet ↔ a0 ∈ s0 \ x0.toSet

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2
Proof step 8:
⊦ a ∈ x.toSet ↔ a ∈ x

Rule argument

spec: set-union name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [x0.toSet, s0 ++ a, a0] flags: <all paths>

Additional information

Used simplifier rules: toSet-ext: ⊦ a ∈ x.toSet ↔ a ∈ x;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (ws) Used terms: x0.toSet, s0 ++ a, a0