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Proof step 40 from proof uni-ass (previous step)

H ⊥ H0, H1 ⊥ H, H0 ⊥ H1 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: ext antecedent: succedent: H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) substlist: [H1, H0] → [H0 ∪ H1 ∪ H, (H0 ∪ H1) ∪ H] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((1, 0, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (H0 ∪ H1 ∪ H, (H0 ∪ H1) ∪ H)]]