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Proof step 1 from proof at-explicit-other (conclusion)
m ≠ n ⊦ @(m, m0, m1) ∈ st → st[n, ablock][@(m, m0, m1)] = st[@(m, m0, m1)]
System applied rule normalize
And obtained the following new goals
Proof step 2
:
@(m, m0, m1) ∈ st ⊦ st[n, ablock][@(m, m0, m1)] = st[@(m, m0, m1)], m = n
Rule argument
Additional information