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Proof step 10 from proof eq-shift-del (previous step)

a0 ∈ st ↔ a0 ∈ st0, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0]), a0 ∈ st0 → st[a0] = st0[a0] ⊦ a0 = a, a ∈ st, a ∈ st0 ∧ st0[a] = d

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a abort: F discard: F computed: T

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%At-same: ⊦ st[a, d][a] = d; Used terms: a