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Proof step 17 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# 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))

Rule argument

Additional information

Used simplifier rules: append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)