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Proof step 27 from proof size-two (previous step)

s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 29:
∀ a1. a1 ∈ s0 ++ a ↔ a1 ∈ ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ a0, s0 = ∅ ++ b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

spec: <toplevel> name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅ ++ a0 ++ b, s0 ++ a] flags: <all paths>

Additional information

Used terms: ∅ ++ a0 ++ b, s0 ++ a