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

∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0 ↔ (∀ n. ¬ m0 < n → buf[n] = empty)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
(buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0, ∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0 ↔ (∀ n. ¬ m0 < n → buf[n] = empty)
Proof step 12:
∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0 ↔ (∀ n. ¬ m0 < n → buf[n] = empty)

Rule argument

fma: (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0

Additional information

Used formulas: (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0