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Proof step 2 from proof ebm-cons-lebsize (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ v ∈ avols ∧ m < # avols[v] ∧ avols[v][m] = mapped(buf, n) ∧ ebm-cons(avols) → # buf = LEB_SIZE; (s)