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Proof step 5 from proof restrict-add-subseteq-at (previous step)

adr ∈ iast0, ∀ adr0. adr0 ∈ iast0 | (as ++ adr) → adr0 ∈ iast1 ∧ (iast0 | (as ++ adr))[adr0] = iast1[adr0] ⊦ iast1[adr] = iast0[adr]

User applied rule all left

And obtained the following new goals

Rule argument

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

Additional information

Used terms: adr