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Proof step 2 from proof ebm-cons-above-fillcount (previous step)

avols[v][m] = mapped(buf, m0), m0 + m1 < LEB_SIZE, m < # avols[v], ebm-cons(avols), v ∈ avols ⊦ buf[m1 + m0] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
avols[v][m] = mapped(buf, m0), m0 + m1 < LEB_SIZE, m < # avols[v], ebm-cons(avols), v ∈ avols ⊦ buf[m1 + m0] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ LEB_SIZE ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic