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Proof step 2 from proof subset-eqval (previous step)

∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], rk ∈ st ⊦ st0[rk] .intval = st[rk] .intval

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (rk)abort:F, discard:T, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk)]]