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Proof step 3 from proof onlymod-shift (previous step)

(((r0 = r1 ∧ sk = sk0) ⊃ val; st[r0 -- sk])) = st1[r0 -- sk], r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, r1 ∈ st, r1 ∈ st1, ∀ r2, sk1. r2 ≠ r → (((r2 = r1 ∧ sk1 = sk0) ⊃ val; st[r2 -- sk1])) = st1[r2 -- sk1], r -- _type ∈ st → (((r = r1 ∧ _type = sk0) ⊃ val; st[r -- _type])) = st1[r -- _type], r -- _length ∈ st → (((r = r1 ∧ _length = sk0) ⊃ val; st[r -- _length])) = st1[r -- _length], ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1 ⊦ st[r0 -- sk] = (((r0 = r1 ∧ sk = sk0) ⊃ st[r1 -- sk0]; st1[r0 -- sk])), r0 = r

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]