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Proof step 1 from proof cons-eq-inf (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ⌜nf⌝ = ⌜a0 '⌝ + σ ↔ nf(0) = a0 ∧ ⌜λ n. nf(n + 1)⌝ = σ

Rule argument

Additional information

Used simplifier rules: cons-append: ⊦ a + σ = ⌜a '⌝ + σ;