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Proof step 12 from proof isempty-elim-left (previous step)

n0 < # buf1, n < # buf1, n < n0 + m, n0 ≤ n, n0 + m ≤ # buf1, ∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty ⊦ buf1[n] = empty, m = 0

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n + # buf0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # buf ⊦ (buf0 + buf)[# buf0 + n] = buf[n]; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer] Used terms: n + # buf0