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Proof step 12 from proof subset-put-02 (
previous step
)
a ∈ st0, a0 ∈ st, ∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ st[a0] = st0[a0] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]
User applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: a0 abort: F discard: F computed: T
Additional information
Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used terms: a0