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Proof step 12 from proof eq-shift-put (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

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