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Proof step 18 from proof is-encoded-prefix (previous step)

# buf0 + # buf1 = # buf2 + # buf3, n < # buf0, n < # buf2, is-encoded(a, buf0), # buf0 ≤ # buf2, ∀ n0. n0 < # buf2 + # buf3 → (buf0 + buf1)[n0] = (buf2 + buf3)[n0] ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, buf2[n] = buf0[n], is-encoded(a, subarray(buf2, 0, # buf0))

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ n < # buf → (buf + buf0)[n] = buf[n]; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%# buf = n ⊦ subarray(buf, 0, n) = buf; (s) Used terms: n