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Proof step 3 from proof sub-size-eq (
previous step
)
# s0 ≤ # s, s ⊆ s0 ⊦ s = s0
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 4
:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 5
:
# s0 ≤ # s, s ⊆ s0 ⊦ ∀ a. a ∈ s ↔ a ∈ s0
Rule argument
spec: <toplevel> name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [s0, s] flags: <all paths>
Additional information
Used terms: s0, s