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Proof step 19 from proof uni-inj (conclusion)

⊦ H ⊥ H0 ∧ H ⊥ H1 → (H ∪ H0 = H ∪ H1 ↔ H0 = H1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
H ∪ H0 = H ∪ H1, H ⊥ H0, H ⊥ H1 ⊦ H0 = H1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]