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Proof step 6 from proof eq-shift-del (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 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;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used terms: a