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Proof step 1 from proof extensionality-to (conclusion)
⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ¬ # σ0 < ⌜n⌝ → σ0 to n = σ1 to n)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
# σ0 = # σ1, ∀ n. ¬ # σ1 < ⌜n⌝ → σ0 to n = σ1 to n ⊦ σ0 = σ1
Rule argument
Additional information