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Proof step 4 from proof in-diffs (previous step)

∀ r, H. r ∈ H \ rs0 ↔ r ∈ H ∧ ¬ r ∈ rs0 ⊦ r ∈ (H -- r0) \ rs0 ↔ r ∈ H ∧ r ≠ r0 ∧ ¬ r ∈ rs0

User applied rule all left

And obtained the following new goals

Proof step 5:
r ∈ (H -- r0) \ rs0 ↔ r ∈ H -- r0 ∧ ¬ r ∈ rs0, ∀ r, H. r ∈ H \ rs0 ↔ r ∈ H ∧ ¬ r ∈ rs0 ⊦ r ∈ (H -- r0) \ rs0 ↔ r ∈ H ∧ r ≠ r0 ∧ ¬ r ∈ rs0

Rule argument

pos: left 1 terms: r, H -- r0 abort: F discard: F computed: T

Additional information

Used terms: r, H -- r0