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Proof step 8 from proof subsize-eq (previous step)

# s = # s0, a ∈ s0, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 10:
# s = # s0, ∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s

Rule argument

spec: inst: name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s0, a] flags: rewrite, current

Additional information

Used terms: s0, a