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Proof step 6 from proof subseteq-rem-del (previous step)

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

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

Rule argument

spec: <toplevel> name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [s0 -- a, s1, a0] flags: <all paths>

Additional information

Used terms: s0 -- a, s1, a0