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Proof step 27 from proof eqex-tput (previous step)

eqex(st1, st0[r -- sk, val], r0), eqex(st0[r -- sk, val], st1, r0), eqex(st, st0, r0), eqex(st0, st, r0), r1 -- sk0 ∈ st, r1 ∈ st, ∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk], ∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r0 ↔ rk0 ∈ st1 -- r0) ∧ (st0[r -- sk, val] -- r0)[rk0] = (st1 -- r0)[rk0] ⊦ r1 = r, r0 = r1, r1 -- sk0 ∈ st1, r1 -- sk0 ∈ st0

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 7, [extquanttermlist: (r1 -- sk0) abort:F , discard:F , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st])], [EXTRATERMS : (r1 -- sk0)]]