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Proof step 1 from proof Data-notin-other-3 (conclusion)
⊦ d ≠ d0 ∧ d0 ≠ d1 ∧ d0 ≠ d2 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1][b, d2][c, d]
System applied rule normalize
And obtained the following new goals
Proof step 2
:
d0 ∈ st[a, d1][b, d2][c, d] ⊦ d = d0, d0 = d2, d0 = d1, d0 ∈ st
Rule argument
Additional information