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Proof step 1 from proof get-cons-num (conclusion)
n ≠ 0 ⊦ (⌜a '⌝ + σ)[(* n)] = σ[n - 1]
Special step normalize
And obtained the following new goals
Proof step 2
:
n ≠ 0 ⊦ (⌜a '⌝ + σ)[n] = σ[n - 1]