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Proof step 10 from proof subseteq-rem-del (previous step)

a ∈ s0, s0 ⊆ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ 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 11:
⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)
Proof step 12:
a ∈ s0, ∀ a. a ∈ s0 → a ∈ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ s0 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

Rule argument

spec: set-basic inst: name: Subset seq: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2) subst: [s2, s1] → [s1, s0] flags: rewrite

Additional information

Used terms: s1, s0