Commutative Property Variation 1

Substitution 2

Substitution 6

Zero Plus a

Addi to Zero

Transitive Property of Equality Variation 2

Addi Zero Insn 0

Transitive Property Application 2

Pc 0

Addi Zero Insn 1

Pc 1

Byte 3 Stays the Same 1

Addi Zero Insn 2

Pc 2

Byte 3 Stays the Same 2

Byte 4 Stays the Same 2

Addi Insn 3

Pc 3

Byte 3 Stays the Same 3

Byte 4 Stays the Same 3

Addi Insn 4

Pc 4

Byte 3 Stays the Same 4

Byte 5 Stays the Same 4

Truth Propagation Property 2

Beq No Branch 5

Byte 3 Stays the Same 5

Byte 4 Stays the Same 5

Byte 5 Stays the Same 5

Jump Help 6

Byte 3 Stays the Same 6

Byte 4 Stays the Same 6

Byte 5 Stays the Same 6

Addi Insn 7

Pc 7

Byte 3 Stays the Same 7

Byte 4 Stays the Same 7

Addi Insn 8

Pc 8

Byte 3 Stays the Same 8

Byte 5 Stays the Same 8

Beq No Branch 9

Byte 3 Stays the Same 9

Byte 4 Stays the Same 9

Byte 5 Stays the Same 9

Jump Help 10

Byte 3 Stays the Same 10

Byte 4 Stays the Same 10

Byte 5 Stays the Same 10

Addi Insn 11

Pc 11

Byte 3 Stays the Same 11

Byte 4 Stays the Same 11

Addi Insn 12

Pc 12

Byte 3 Stays the Same 12

Byte 5 Stays the Same 12

Transitive Property of Equality Variation 1

Beq Branch

Byte 3 Stays the Same 13

Byte 4 Stays the Same 13

Byte 5 Stays the Same 13

Loop

Proof: Pc 12

Let's prove the following theorem:

if the following are true:

instruction #4 is addi dst=4 src=4 imm=1
the PC at time 12 = 4

then the PC at time 13 = 5

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	instruction #4 is `addi dst=4 src=4 imm=1`
2	the PC at time 12 = 4

Proof Table

#	Claim	Reason
1	the PC at time (12 + 1) = 4 + 1	if instruction #4 is `addi dst=4 src=4 imm=1` and the PC at time 12 = 4, then the PC at time (12 + 1) = 4 + 1 (Program Counter on ADDI)
2	12 + 1 = 13	12 + 1 = 13 (Fixed Value Statement)
3	4 + 1 = 5	4 + 1 = 5 (Fixed Value Statement)
4	the PC at time (12 + 1) = the PC at time 13	if 12 + 1 = 13, then the PC at time (12 + 1) = the PC at time 13 (Substitution)
5	the PC at time 13 = 5	if the PC at time (12 + 1) = 4 + 1 and the PC at time (12 + 1) = the PC at time 13 and 4 + 1 = 5, then the PC at time 13 = 5 (Transitive Property Application 2)

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