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Proof step 4 from proof datasize-ls (previous step)

⊦ datasize(buf, # buf) < m ↔ ¬(∃ n. m ≤ n + 1 ∧ n < # buf ∧ buf[n] ≠ empty), m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ m0 ≤ # buf ∧ m ≠ 0 → (datasize(buf, m0) < m ↔ ¬(∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty))
Proof step 6:
# buf ≤ # buf ⊦ ¬(∃ n. n + 1 ≥ m ∧ n < # buf ∧ buf[n] ≠ empty) ↔ ¬(∃ n. m ≤ n + 1 ∧ n < # buf ∧ buf[n] ≠ empty), m = 0

Rule argument

spec: <toplevel> name: datasize-ls-h seq: ⊦ m0 ≤ # buf ∧ m ≠ 0 → ( datasize(buf, m0) < m ↔ ¬(∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty)) subst: [m, m0, buf] → [m, # buf, buf] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) Used terms: m, # buf, buf