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Proof step 4 from proof sub-trans (
previous step
)
st[a] = st0[a], a ∈ st, a ∈ st0, ∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a], ∀ a. a ∈ st0 → a ∈ st1 ∧ st0[a] = st1[a] ⊦ a ∈ st1 ∧ st0[a] = st1[a]
User applied rule all left
And closed the goal!
Rule argument
pos: left 5 terms: a abort: F discard: F computed: T
Additional information
Used terms: a