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Proof step 2 from proof k-diff-rec1 (previous step)

a ∈ s0, a1 ∈ s0 ⊦ a0 ∈ s \ s0 ↔ a0 ∈ s \ (s0 -- a), a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2

Rule argument

spec: inst: name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [s0, s, a0] flags: rewrite, current

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s); Used terms: s0, s, a0