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Proof step 11 from proof datasize-empbuf-h (previous step)

m0 < # buf + n, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ ((buf + mkempbuf(n))[m0] = empty ⊃ datasize(buf + mkempbuf(n), m0);m0 + 1) = datasize(buf, (¬ m0 < # buf ⊃ # buf;m0 + 1))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 12:
(buf + mkempbuf(n))[m0] = empty, m0 < # buf + n, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ datasize(buf + mkempbuf(n), m0) = datasize(buf, (¬ m0 < # buf ⊃ # buf;m0 + 1))
Proof step 22:
m0 < # buf + n, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ m0 + 1 = datasize(buf, (¬ m0 < # buf ⊃ # buf;m0 + 1)), (buf + mkempbuf(n))[m0] = empty

Rule argument

pos: right 1

Proof step 11 from proof datasize-empbuf-h (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

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