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Proof step 31 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
buf0 + buf1 = buf2 + buf3, n < # buf0, n < # buf2, # buf2 < # buf0, n + 1 < # buf0, is-encoded(a, buf0), is-encoded(b, buf2) ⊦ # buf0 = # buf2, buf0[n] = buf2[n], is-encoded(b, subarray(buf0, 0, # buf2))

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%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)