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Proof step 8 from proof ext-rec (
previous step
)
a ∈ st1, ∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b]) ⊦ a0 = a, (a0 ∈ st1 ↔ a0 ∈ st0) ∧ st1[a0] = st0[a0]
User applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a0 abort: F discard: F computed: T
Additional information
Used simplifier rules: notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a]; Used terms: a0