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Proof step 34 from proof size-two (previous step)

∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0, b = a ∨ b ∈ s0 ⊦ 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 35:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 36:
∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, b = a ∨ b ∈ s0, a0 = a ∨ a0 ∈ s0 ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ a0, a ∈ s0, ∀ a. a ∈ s0 ↔ a ∈ ∅ ++ b, ∃ a. s0 = ∅ ++ a

Rule argument

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

Additional information

Used terms: ∅ ++ b, s0