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Proof step 18 from proof from-succ (previous step)

⊦ ⌜nf from n⌝ = ⌜λ n0. (n0 = 0 ⊃ nf(n);nf((n + 1) + n0 -1))⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ nf(n0 + n) = (n0 = 0 ⊃ nf(n);nf((n + 1) + (n0 - 1)))

Rule argument

Additional information

Used simplifier rules: get-from-inf: ⊦ ((nf from n))(m) = nf(m + n);% ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)% ⊦ n -1 = n - 1; (s)