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Proof step 1 from proof abs-of-emp (conclusion)

⊦ abs-of(∅) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ofh = ofh0 ↔ (∀ fd. (fd ∈ ofh ↔ fd ∈ ofh0) ∧ ofh[fd] = ofh0[fd])
Proof step 3:
⊦ ∀ fd. (fd ∈ abs-of(∅) ↔ fd ∈ ∅) ∧ abs-of(∅)[fd] = (∅)[fd]

Rule argument

spec: store, inst: open-file-handles name: Extension seq: ⊦ ofh = ofh0 ↔ (∀ fd. (fd ∈ ofh ↔ fd ∈ ofh0) ∧ ofh[fd] = ofh0[fd]) subst: [ofh0, ofh] → [∅, abs-of(∅)] flags: <all paths>

Additional information

Used terms: ∅, abs-of(∅)