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Proof step 3 from proof cons-eq-inf (previous step)

⊦ ⌜nf⌝ = ⌜a0 '⌝ + ⌜nf0⌝ ↔ nf(0) = a0 ∧ ⌜λ n. nf(n + 1)⌝ = ⌜nf0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ⌜nf⌝ = ⌜a0 '⌝ + ⌜nf0⌝ ↔ nf(0) = a0 ∧ (λ n. nf(n + 1)) = nf0

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)