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Proof step 14 from proof eq-shift-del (previous step)

a ∈ st0, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ⊦ a0 = a, a ∈ st, (a0 ∈ st ↔ a0 ∈ st0) ∧ (a0 ∈ st → st[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: Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a];%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used terms: a0