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Proof step 61 from proof datasize-further (previous step)

datasize(buf, n) = datasize(buf, n0) ↔ (∀ n1. n ≤ n1 ∧ ¬ n0 < n1 → buf[n1] = empty), buf[n0] = empty, n < # buf, n ≤ n0, ∀ buf, n. (datasize(buf, n) = datasize(buf, n0) ↔ (∀ n2. n ≤ n2 ∧ n2 < n0 → buf[n2] = empty)) ∧ n < n0 ∧ n < # buf ∧ (∀ n2. n < n2 ∧ n2 < n0 → buf[n2] = empty) → n + 1 = datasize(buf, n0) ∨ buf[n] = empty, ∀ n1. n < n1 ∧ ¬ n0 < n1 → buf[n1] = empty ⊦ n + 1 = datasize(buf, n0), buf[n] = empty

User applied rule all left

And obtained the following new goals

Proof step 62:
datasize(buf, n) = datasize(buf, n0) ↔ (∀ n1. n ≤ n1 ∧ ¬ n0 < n1 → buf[n1] = empty), buf[n0] = empty, n < # buf, n ≤ n0, (datasize(buf, n) = datasize(buf, n0) ↔ (∀ n2. n ≤ n2 ∧ n2 < n0 → buf[n2] = empty)) ∧ n < n0 ∧ n < # buf ∧ (∀ n2. n < n2 ∧ n2 < n0 → buf[n2] = empty) → n + 1 = datasize(buf, n0) ∨ buf[n] = empty, ∀ n1. n < n1 ∧ ¬ n0 < n1 → buf[n1] = empty ⊦ n + 1 = datasize(buf, n0), buf[n] = empty

Rule argument

pos: left 5 terms: buf, n abort: F discard: T computed: T

Additional information

Used terms: buf, n