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Proof step 44 from proof precise-sep (previous step)

H1 ∪ H0 = H3 ∪ H, P0(H4), P1(H1), P1(H3), H4 ⊥ H3 ∪ H, H0 ⊥ H1, H1 ⊥ H4, H ⊥ H3, H4 ⊥ H, H0 ⊥ H4, H4 ⊥ H3, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H1 = H3

User applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: H1, H3, H0, H abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H1, H3, H0, H