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Proof step 26 from proof recovery (previous step)

ebm-cons(avols0), ebm-cons(avols1), ebm-vol-inv(abs(avols1[default-volid])), default-volid ∈ avols1, avols1 ⊆ avols0 ⊦ abs(avols1[default-volid]) ⊆ abs(avols0[default-volid])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
ebm-cons(avols1), ebm-cons(avols0), ebm-vol-inv(abs(avols1[default-volid])), default-volid ∈ avols1, avols1 ⊆ avols0 ⊦ abs(avols1[default-volid]) ⊆ abs(avols0[default-volid])

Proof step 26 from proof recovery (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information