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Proof step 9 from proof eq-union-disj (previous step)

zh1 ⊥ zh, zh0 ⊥ zh, ∀ r0. (¬ r0 ∈ zh → (r0 ∈ zh0 ↔ r0 ∈ zh1)) ∧ (r0 ∈ zh0 ∨ r0 ∈ zh → (r0 ∈ zh0 ⊃ zh0[r0];zh[r0]) = (zh1 ∪ zh)[r0]) ⊦ (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])

User applied rule all left

And obtained the following new goals

Proof step 10:
zh0 ⊥ zh, zh1 ⊥ zh, ¬ r ∈ zh → (r ∈ zh0 ↔ r ∈ zh1), r ∈ zh0 ∨ r ∈ zh → (r ∈ zh0 ⊃ zh0[r];zh[r]) = (zh1 ∪ zh)[r] ⊦ (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])

Rule argument

pos: left 3 terms: r abort: F discard: T computed: T

Additional information

Used terms: r