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Proof step 51 from proof eqex-tput (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 50:
⊦ eqex(st1, st0[r -- sk, val], r0) ∧ eqex(st0, st, r0) ∧ eqex(st0[r -- sk, val], st1, r0) ∧ eqex(st, st0, r0) ∧ (∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r0 ↔ rk0 ∈ st1 -- r0) ∧ (st0[r -- sk, val] -- r0)[rk0] = (st1 -- r0)[rk0]) ∧ (∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk]) → (st[r -- sk, val] -- r0)[r1 -- sk0] = (st1 -- r0)[r1 -- sk0] ∧ (r1 ≠ r0 ∧ r1 -- sk0 ∈ st[r -- sk, val] ↔ r1 -- sk0 ∈ st1 -- r0)

Rule argument

[VARWITHFMAVARSARG : rk, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : refkey generated by --;])]]