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Proof step 48 from proof eqex-tput (previous step)

r0 ≠ r1, 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][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 47:
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] ⊦ r0 = r1, (r1 -- sk0 ∈ st[r -- sk, val] ↔ r1 -- sk0 ∈ st1) ∧ (((r1 = r ∧ sk0 = sk) ⊃ val; st[r1 -- sk0])) = st1[r1 -- sk0]

Rule argument

[emptyarg]

Additional information

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