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Proof step 14 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
buf0 + buf1 = buf2 + buf3, n < # buf0, n < # buf2, is-encoded(a, buf0), # buf0 ≤ # buf2 ⊦ 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: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(buf, m, n)[n0] = buf[m + n0]; (s)% ⊦ # subarray(buf, m, n) = n; (s)