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Proof step 9 from proof subset-uni (previous step)

H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0 ⊦ ((H1 \ H0) ∪ H0)[r] = H1[r]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])
Proof step 11:
r ∈ (H1 \ H0) ∪ H0, H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0 ⊦ (r ∈ H1 \ H0 ⊃ (H1 \ H0)[r];H0[r]) = H1[r]

Rule argument

spec: <toplevel> name: at-uni seq: r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]) subst: [r, H1, H0] → [r, H0, H1 \ H0] flags: <all paths>

Additional information

Used simplifier rules: in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);%diff-in: ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1; Used terms: r, H0, H1 \ H0