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Proof step 50 from proof eqex-tput (previous step)

⊦ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ eqex(st, st0, r) ∧ eqex(st0, st1, r) → eqex(st, st1, r)
Proof step 14:
⊦ eqex(st, st0, r) → eqex(st0, st, r)
Proof step 15:
⊦ eqex(st, st, r)
Proof step 49:
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] ⊦ (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

[emptyarg]

Additional information

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