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Proof step 2 from proof avol-down-mapped-eq (previous step)

avol0[n] = mapped(buf), n < # avol0, ebm-vol-inv(avol0), avol0 ⊆ avol1 ⊦ avol1[n] = mapped(buf)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
avol0[n] = mapped(buf), n < # avol0, ebm-vol-inv(avol0), avol0 ⊆ avol1 ⊦ avol1[n] = mapped(buf)

Rule argument

Additional information

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