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Proof step 36 from proof eqex-tput (previous step)

r1 = r ∧ sk0 = sk ∨ r1 -- sk0 ∈ st0 ↔ r1 -- sk0 ∈ st1, (((r1 = r ∧ sk0 = sk) ⊃ val; st0[r1 -- sk0])) = st1[r1 -- sk0], 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 ∨ r1 = r ∧ sk0 = sk ↔ r1 -- sk0 ∈ st1

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 35:
r1 ≠ r ∨ sk0 ≠ sk, st0[r1 -- sk0] = st1[r1 -- sk0], r1 = r ∧ sk0 = sk ∨ r1 -- sk0 ∈ st0 ↔ r1 -- sk0 ∈ st1, 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 ∨ r1 = r ∧ sk0 = sk ↔ r1 -- sk0 ∈ st1

Rule argument

[CASEDARG : pos: left 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]