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Proof step 5 from proof size-uni (previous step)

H0 ⊥ H1, ∀ H, H0. # H < # H1 → H0 ⊥ H → #(H0 ∪ H) = # H0 + # H ⊦ #(H0 ∪ H1) = # H0 + # H1, H1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 7:
H0 ⊥ H1, ∀ H, H0. # H < # H1 → H0 ⊥ H → #(H0 ∪ H) = # H0 + # H ⊦ #(H0 ∪ H1) = # H0 + # H1, ∀ r. (r ∈ H1 ↔ r ∈ ∅) ∧ (r ∈ H1 → H1[r] = (∅)[r])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [∅, H1] flags: <all paths>

Additional information

Used terms: ∅, H1