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Proof step 7 from proof isempty-elim-left0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m ≤ # buf1 ⊦ (∀ n. # buf0 ≤ n ∧ n < # buf0 + m → (buf0 + buf1)[n] = empty) ↔ (∀ n. n < m → buf1[n] = empty)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ 0 ≤ m; (s)