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Proof step 8 from proof subseteq-rem-del (previous step)

a ∈ s0, s0 ⊆ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 \ s0 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2
Proof step 10:
a ∈ s0, s0 ⊆ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ s0 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

Rule argument

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

Additional information

Used terms: s0, s1, a0