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Proof step 6 from proof bmin-sub-01 (previous step)

# s = 2, s ⊆ s0 ⊦ s.butmin.min ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)
Proof step 8:
s ⊆ s0, ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ⊦ s.butmin.min ∈ s0

Rule argument

spec: <toplevel> name: size-two seq: ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b) subst: [s] → [s] flags: <all paths>

Additional information

Used terms: s