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Proof step 5 from proof datasize-datasize-elim-h (previous step)

n0 < # buf, ∀ buf. n0 ≤ # buf → datasize(buf, datasize(buf, n0)) = datasize(buf, n0) ⊦ datasize(buf, (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1)) = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
buf[n0] = empty, n0 < # buf, ∀ buf. n0 ≤ # buf → datasize(buf, datasize(buf, n0)) = datasize(buf, n0) ⊦ datasize(buf, datasize(buf, n0)) = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1)
Proof step 8:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)

Rule argument

pos: right 1

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)