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Proof step 3 from proof sizesubl (previous step)

a ∈ s0, # s0 ≤ # s, s ⊆ s0 ⊦ a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 5:
# s0 ≤ # s, s ⊆ s0, ∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1 ⊦ a ∈ s

Rule argument

spec: <toplevel> name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s0, a] flags: <all paths>

Additional information

Used terms: s0, a