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Proof step 41 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), ∀ 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] ⊦ val = st1[r -- sk], r0 = r

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [QUANTTERMLIST : (r -- sk)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st], [CSIMPSEQ : ⊦ rk ∈ st[rk, val]], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ r ≠ r0 → (st -- r)[r0 -- sk0] = st[r0 -- sk0]], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))])], [EXTRATERMS : (r -- sk)]]