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Proof step 19 from proof from-succ (previous step)

⊦ nf(n0 + n) = (n0 = 0 ⊃ nf(n);nf((n + 1) + (n0 - 1)))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 20:
⊦ n0 = 0, nf(n0 + n) = nf((n + 1) + (n0 - 1))

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic