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Proof step 6 from proof at-uni (previous step)

H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1, ∀ H, H0, r. # H < # H1 → H0 ⊥ H ∧ (r ∈ H0 ∨ r ∈ H) → (H0 ∪ H)[r] = (r ∈ H0 ⊃ H0[r];H[r]) ⊦ H1 = ∅, (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 8:
H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1, ∀ H, H0, r. # H < # H1 → H0 ⊥ H ∧ (r ∈ H0 ∨ r ∈ H) → (H0 ∪ H)[r] = (r ∈ H0 ⊃ H0[r];H[r]) ⊦ (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]), ∀ 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