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Proof step 9 from proof subset-put-01 (previous step)

a1 ∈ st, a ∈ st0[a, st[a]], a ∈ st, ∀ a0. a0 ∈ st → a0 ∈ st0[a, st[a]] ∧ st[a0] = st0[a, st[a]][a0] ⊦ a1 = a, a1 ∈ st0 ∧ st[a1] = st0[a1]

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a1 abort: F discard: F computed: T

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; Used terms: a1