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Proof step 11 from proof nodes-no-change-eq (previous step)

m ∈ gstore0 ↔ m ∈ gstore1, gstore1[m] = groupblock(gndx, lax, m2), m ∈ gstore1 → gndx = gstore0[m].gndlist ∧ lax = gstore0[m].ladrlist ⊦ (¬ m ∈ ginval → (@(m, m0, m1) ∈ gstore0 ↔ @(m, m0, m1) ∈ gstore1)) ∧ (@(m, m0, m1) ∈ gstore0 ∧ ¬ m ∈ ginval → nodes(gstore0, ginval)[@(m, m0, m1)] = nodes(gstore1, ginval)[@(m, m0, m1)])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
m ∈ gstore0 ↔ m ∈ gstore1, gstore1[m] = groupblock(gndx, lax, m2), gstore0[m] = groupblock(gndx0, lax0, m3), m ∈ gstore1 → gndx = gndx0 ∧ lax = lax0 ⊦ (¬ m ∈ ginval → (@(m, m0, m1) ∈ gstore0 ↔ @(m, m0, m1) ∈ gstore1)) ∧ (@(m, m0, m1) ∈ gstore0 ∧ ¬ m ∈ ginval → nodes(gstore0, ginval)[@(m, m0, m1)] = nodes(gstore1, ginval)[@(m, m0, m1)])

Rule argument

spec: ablock, inst: group-store name: elim seq: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m) subst: [gblock, gndx, lax, m] → [gstore0[m], gndx0, lax0, m3]

Additional information

Used simplifier rules: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m); (elim) Used terms: gstore0[m], gndx0, lax0, m3