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Proof step 35 from proof is-encoded-prefix (previous step)

# buf0 + # buf1 = # buf2 + # buf3, n < # buf0, n < # buf2, # buf2 < # buf0, n + 1 < # buf0, is-encoded(a, buf0), is-encoded(b, buf2), ∀ n0. n0 < # buf2 + # buf3 → (buf0 + buf1)[n0] = (buf2 + buf3)[n0] ⊦ # buf0 = # buf2, buf0[n] = buf2[n], is-encoded(b, subarray(buf0, 0, # buf2))

User applied rule all left

And closed the goal!

Rule argument

pos: left 8 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 ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n